how to implement Bessel's function in MATLAB

i want to find the solution of this bessels function mentioned

please help me to model the same in matlab or simulink

Answer 1

[Walter Roberson]

https://www.mathworks.com/help/matlab/ref/besselj.html for the numeric version

https://www.mathworks.com/help/symbolic/besselj.html for the symbolic version

When Bessel functions are being used, it is not uncommon to find that you need higher precision than double precision can provide, so it is common to need to use the symbolic version. To be more precise, often calculation of the values needs a higher range than double precision can support: it is common to end up with formulas that involve the ratio of two large numbers, and even though the ratio might be in the 1/1000 to 2 range, the individual numbers overflow to infinity in double precision.

[NIHAD MOHAMED ALI]

could you be more specific, on how to implement that equation in matlab

[Walter Roberson]

is besselj(1, sigma1)

[NIHAD MOHAMED ALI]

how do i implement J1'(sigma) ?

[Walter Roberson]

syms sigma
J1prime(sigma) = diff(besselj(1,sigma))

J1prime(sigma) = 

vpa(J1prime(3))

ans = 

[NIHAD MOHAMED ALI]

Thanks a lot sir,

[NIHAD MOHAMED ALI]

sigma=1.469;
x=besselj(1,sigma);
y=bessely(1,sigma);
z=diff(besselj(1,sigma));
vpa(z);

what am i doing wrong?
am geting answer as zero

[NIHAD MOHAMED ALI]

@Walter Roberson specific thing is i need to solve equation 2 and use the result in equation 3! so am studying the basics you been of great help,but iam still having trouble cracking it

[Walter Roberson]

format long g
sigma=1.469

sigma = 
                     1.469

x = besselj(1,sigma)

x = 
         0.553407191636554

y = bessely(1,sigma)

y = 
        -0.432787705894754

syms Sigma
J1prime(Sigma) = diff(besselj(1,Sigma))

J1prime(Sigma) = 

z = J1prime(sigma)

z = 

vpa(z)

ans = 

0.15233074502500402865842627859807

[Walter Roberson]

Remember, besselj(1,sigma) is a constant value, and diff() of a constant value is 0. You did something similar to

syms X
f(X) = X^2

f(X) = 

x = 3

x = 3

diff(f(x))

ans = 

0

df = diff(f)

df(X) = 

df(x)

ans = 

6

When x is a particular number, f(x) is a point, not a line, and diff() of a point is 0. You need to differentiate the function and then substitute the particular number into the result.

[NIHAD MOHAMED ALI]

yes, clealy understood my mistake ; thanks a bunch