Is there any one who can help me how to sum the following summation expression? I tried but it said to many inputs. Thank you for your help!

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Temesgen Gelaw
Temesgen Gelaw 2016-6-11
评论： Temesgen Gelaw 2016-6-12
U(x,y) = ∑[∝.ϕ1(x) + β.ϕ2(x)].ψ(y), the summation is k : 0 --> infinity.
Where ψ(y)=sin(kπy), b=√(1/(2ϵ)^2 +(kπ)^2 ),
ϕ1(x)=e^(x/(2ϵ)) (sinh(b.(1-x)))/(sinh(b)),
ϕ2(x)=e^((x-1)/(2ϵ)).(sinh(b.x)/(sinh(b)),
∝=2/kπ (cos(0.45kπ) ),
β=∝.(b.e^(1/(2ϵ))/(sinh(b))⁄(1/2ϵ+b.(cosh(b)/sinh(b)))
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Temesgen Gelaw
Temesgen Gelaw 2016-6-11
编辑：Temesgen Gelaw 2016-6-11
Dear Geoff
First, I write like this
syms x y k v
Hk = sinh(k*pi*y);
bk = sqrt(1/(2*v).^2 + (k*pi).^2);
Qlk = exp(x/(2*v))*sinh(bk*(1-x))/sinh(bk);
Q2k = exp((x-1)/(2*v))*sinh(bk*x)/sinh(bk);
Ck = 2*(cos(0.45*k*pi)-cos(0.55*k*pi))/(k*pi);
Bk = Ck*((bk*exp(1/(2*v))/sinh(bk)))/(1/(2*v) + bk*cosh(bk)/sinh(bk));
U = (Ck*Qlk + Bk*Q2k)*Hk;
u = symsum(U, k, 1, Inf);
Then I got this
u = -sum((exp(pi*k*y)/2 - exp(-pi*k*y)/2)*((exp(x/(2*v))*(exp((1/(4*v^2) + pi^2*k^2)^(1/2)*(x - 1))/2 - exp(-(1/(4*v^2) + pi^2*k^2)^(1/2)*(x - 1))/2)*(exp(-(pi*k*9*i)/20) + exp((pi*k*9*i)/20) - exp(-(pi*k*11*i)/20) - exp((pi*k*11*i)/20)))/(pi*k*(exp((1/(4*v^2) + pi^2*k^2)^(1/2))/2 - exp(-(1/(4*v^2) + pi^2*k^2)^(1/2))/2)) - (exp((x - 1)/(2*v))*exp(1/(2*v))*(exp(x*(1/(4*v^2) + pi^2*k^2)^(1/2))/2 - exp(-x*(1/(4*v^2) + pi^2*k^2)^(1/2))/2)*(1/(4*v^2) + pi^2*k^2)^(1/2)*(exp(-(pi*k*9*i)/20) + exp((pi*k*9*i)/20) - exp(-(pi*k*11*i)/20) - exp((pi*k*11*i)/20)))/(pi*k*(exp((1/(4*v^2) + pi^2*k^2)^(1/2))/2 - exp(-(1/(4*v^2) + pi^2*k^2)^(1/2))/2)^2*(1/(2*v) + ((1/(4*v^2) + pi^2*k^2)^(1/2)*(exp((1/(4*v^2) + pi^2*k^2)^(1/2))/2 + exp(-(1/(4*v^2) + pi^2*k^2)^(1/2))/2))/(exp((1/(4*v^2) + pi^2*k^2)^(1/2))/2 - exp(-(1/(4*v^2) + pi^2*k^2)^(1/2))/2)))), k == 1..Inf)
And then I tried as below
u = -sum((exp(pi*k*y)/2 - exp(-pi*k*y)/2)*((exp(x/(2*v))*(exp((1/(4*v^2) + pi^2*k^2)^(1/2)*(x - 1))/2 - exp(-(1/(4*v^2) + pi^2*k^2)^(1/2)*(x - 1))/2)*(exp(-(pi*k*9*i)/20) + exp((pi*k*9*i)/20) - exp(-(pi*k*11*i)/20) - exp((pi*k*11*i)/20)))/(pi*k*(exp((1/(4*v^2) + pi^2*k^2)^(1/2))/2 - exp(-(1/(4*v^2) + pi^2*k^2)^(1/2))/2)) - (exp((x - 1)/(2*v))*exp(1/(2*v))*(exp(x*(1/(4*v^2) + pi^2*k^2)^(1/2))/2 - exp(-x*(1/(4*v^2) + pi^2*k^2)^(1/2))/2)*(1/(4*v^2) + pi^2*k^2)^(1/2)*(exp(-(pi*k*9*i)/20) + exp((pi*k*9*i)/20) - exp(-(pi*k*11*i)/20) - exp((pi*k*11*i)/20)))/(pi*k*(exp((1/(4*v^2) + pi^2*k^2)^(1/2))/2 - exp(-(1/(4*v^2) + pi^2*k^2)^(1/2))/2)^2*(1/(2*v) + ((1/(4*v^2) + pi^2*k^2)^(1/2)*(exp((1/(4*v^2) + pi^2*k^2)^(1/2))/2 + exp(-(1/(4*v^2) + pi^2*k^2)^(1/2))/2))/(exp((1/(4*v^2) + pi^2*k^2)^(1/2))/2 - exp(-(1/(4*v^2) + pi^2*k^2)^(1/2))/2)))), k,1,Inf);
But the answer is Too many input arguments.
Image Analyst
Image Analyst 2016-6-11
I don't have the symbolic Toolbox so I can't help you other than to suggest this link: read this
But you must have overlooked Geoff's suggestion " the full error message" since I don't see the error message. You need to give ALL THE RED TEXT including lines numbers, line of code, etc.

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采纳的回答

Roger Stafford
Roger Stafford 2016-6-11
The first result with ‘symsum’ very likely indicates that ‘symsum’ is unable to compute this particular infinite sum.
The second result indicating too many input arguments refers to the ‘sum’ function. See:
http://www.mathworks.com/help/matlab/ref/sum.html
The ‘sum’ function is not symbolic but is numeric instead, and the arguments you gave, namely k,1,Inf, are inappropriate. You need to make ‘k’ a vector and unfortunately it has to be of finite length so you can only approximate an infinite sum with this method.

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