MATLAB: I am trying to get matlab to solve the derivative for t which should come out to 693.15 u. it displays the p value correctly.

errorMATLAB

code:

vnum=input('Enter any numbers before expnential value of V: ');

vcoeff=input('Enter the coeficent to the exponential (V): ');

vexp=input('Enter the exponent to the exponential (V): ');

icoeff=input('Enter the coeficent to the exponential (I): ');

iexp=input('Enter the exponent to the exponential (I): ');

pnum=vnum*icoeff;

pcoeff=vcoeff*icoeff;

pexp=iexp+vexp;

fprintf('p= %0.2fe^( ',pnum)

fprintf('%0.2ft) ',iexp)

fprintf('%0.2fe^(',pcoeff)

fprintf('%0.2ft) W\n',pexp)

fprintf('\n')

syms t;

equation1=(pnum*exp(iexp*t))-pcoeff*exp(pexp*t);

diffequation1=diff(equation1);

pretty(solve(diffequation1,t))

—————————————————————————————————————————

————————————————————————————————————————-

display:

Enter 1 for part A:

Enter 2 for part B:

1

Enter any numbers before expnential value of V: 75

Enter the coeficent to the exponential (V): -75

Enter the exponent to the exponential (V): -1000

Enter the coeficent to the exponential (I): .05

Enter the exponent to the exponential (I): -1000

p= 3.75e^( -1000.00t) -3.75e^(-2000.00t) W

pi 1i log(2)

—– + ——

1000 1000

Best Answer

i have solved the quesion i needed to fprint the answer in sientific notiation and it worked. :)

Related Solutions

MATLAB: Reiteration sequence without Fib function.

Hi Ann,

Sounds like your function should take the following inputs: the first value of your sequence, the second value of your sequence, and the length of the sequence. The function should also return a vector "x" of iterative pair sums.

Be sure that this code is saved in a file name HW215_Ann.m. Then from the Matlab command window, you can type

 x=HW215_Ann(1,2,30);

to call your function (use any inputs you like, i used 1, 2, and 30). The output "x" should then be in your workspace with the sequence you wanted.

Function below:

function x = HW215_Ann(f1,f2,n)
% f1 and f2 are the first two numbers of the sequence
% n is the length of the sequence
% Checking that function inputs are valid
if ~isnumeric(f1)||~isscalar(f1),
   error('First input "f1" must be numeric scalar.');
end
if ~isnumeric(f2)||~isscalar(f2),
   error('Second input "f3" must be numeric scalar.');
end
if ~isnumeric(n)||~isscalar(n),
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elseif (n~=round(n))||n<3,
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% pre-allocate memory for x
x=NaN(n,1);
% start off with f1 and f2
x(1)=f1;
x(2)=f2;
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for iter=3:n
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Functions in Matlab must be saved as .m files with the same name as the as the function's name. Also, you won't need to use return statements in Matlab. The objects you want to return should only be stated in the function declaration. Give this link a read-through.

MATLAB: Fibonacci Sequence and Caluclating a phi value

close, I have tried to keep intact most of what you have done

clear;clc;
num1=input('Enter an integer,greater than three, for the number of terms in the sequence: ');
if (num1<=3 || rem(num1,1)~=0) 
      num1=input('Enter an integer greater than three');
end
num2=input('Enter a integer from 2 to 15 for the precision: ');
if (num2<2 || rem(num2,1)~=0)
      num2=input('Please enter a valid integer between 2 and 15');
end
Npts=num1; m=1; d0=0; d1=1;
while(m<=Npts)
      db(m)=d0;
      d2=d1+d0;
      d0=d1;
      d1=d2;
      m=m+1;
end
disp(db);
iter=0;
tol = 1;
%Npt=1000; f=1; b0=0; b1=1;
bd = db(end-3:end); %create bd
while tol > 10^-num2
%       bd(f)=b0;
%       b2=b1+b0;
%       b0=b1;
%       b1=b2;
%       f=f+1;     you have already created the sequence
        tol = abs((bd(2)/bd(1))-((bd(3)/bd(2))));
%       if(phi< (10^-num2))
%           disp(iter);
%           disp(phi);
%       else
%           t=4;
          bd(1)=bd(2);
          bd(2)=bd(3);
          bd(3)=bd(1)+ bd(2);
%           t=t+1;
%       end
iter=iter+1;
end
fprintf('phi = %1.15f\n',bd(3)./bd(2))
fprintf('iterations = %d\n',iter)

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