Symsum returns symbolic expression instead of value

I would like to solve the equation shown in the picture.

Try to run the code:

syms m 
assume(m,{'positive','integer'})
x=2
y=2
%W=0.4 L=0.3
r=symsum(((1-cos(m*pi))/(m*pi))*sinh((m*pi*(0.4-y)/0.3))/(sinh((m*pi*0.4/0.3)))*sin(m*pi*x/0.3),m,1,inf)

and it returns to:

val =
symsum((sin((20*pi*m)/3)*sinh((16*pi*m)/3)*(cos(pi*m) - 1))/(m*sinh((4*pi*m)/3)), m, 1, Inf)/pi

It would not returns to the exact value, as I tried eval()/vpa().

Not sure what else I can do to solve it.

Thank you for your time and help.

Answer 1

[Steven Lord]

What makes you believe that this summation converges?

syms m
f = sinh(16*pi*m/3)./sinh(4*pi*m/3);
fplot(f, [0 10])

This term takes off like a rocket by the time you reach m = 6 or m = 7.

The fact that you're working with trig functions whose arguments are integer multiples of pi makes me suspect there's a lot of multiplication and division by zero (or would be if pi were exactly πor if you used sinpi and cospi.)

[Walter Roberson]

No division by 0 except at 0. The divisor is sinh which is only 0 at 0.

[Grant Lee]

Sorry, I just noticed that, I plugged in the wrong value for x, and y. Both of them should be 0.0096. But the function still returns to sysmbolic expression.

[Walter Roberson]

With 0.0096 it does appear to converge.

Even m the result is 0 because 1-cos(pi*m) -> 1-1

So you can remap to m = 2*M+1, M=1..infinity .

The number of terms to use depends upon your desired accuracy. It looks to me as if M=160 should get you accuracy to +/- 1e-16

[Grant Lee]

Thank you for your help!