MATLAB: How to calculate roots of the transcendental equation , -K_n*gtan(K_n)= 5, where n= 1,2 ,3,4….g=gavity=9.81

-k_n*gtan(k_n)= 5234find the roots of the transcendental equationg=gavity=9.81where n= 1

Please help to find the roots of the transcendental equation , -K_n*gtan(K_n)= 5, where n= 1,2 ,3,4…., g=gavity=9.81. I have calculte but getting wrong result. Is this is possible to find roots with Newton_Raphson Method

Best Answer

N = 20;
w = sqrt(5);
g = 9.81;
f = @(x) w^2 + g*x.*tan(x);
for i = 1:N
  x0 = [(2*i-1)*pi/2+0.1, (2*i+1)*pi/2-0.1];  
  y(i) = fzero(f,x0);
end 
y

Related Solutions

MATLAB: Sum with odd n+k

I am assuming that "

" is scalar and hence taking it as "r", but that shouldn't matter for this question. Also please initialize r, k before attemting to run the code. Here is the sample code:

sum = 0;
for n = 0:k-1
    if mod(k+n, 2) == 1
       sum = sum + (2*n+1)*r; 
    end
end

Note: This is not an optimized code and you may write a more optimized implementation.

MATLAB: Finding roots of the equation

You may want to check your function, because it doesn't seem to have any (real) zeros:

f = @(n,I) n*210+(I.^2/n).*((((200-32*I).^2)/200)+32)-197;
figure
I = linspace(-100,100,5001);
y = f(1,I);
plot(I,y)
min(y)

But, the approach would be to define a function handle for f, as above, then set the value of n and apply fzero to the resulting function of one variable ( I ):

f = @(n,I) n*210+(I.^2/n).*((((200-32*I).^2)/200)+32)-197;
I0 = 0; % Make an initial guess
for n = 1:5
  g = @(I) f(n,I);  % Make a new function from f(n,I) with n fixed
  I0 = fzero(g,I0)  % Find zero, starting with previous result
end

Best Answer

Related Solutions

MATLAB: Sum with odd n+k

MATLAB: Finding roots of the equation

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