Error while vectorizing my code

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Mohsin Shaikh
Mohsin Shaikh el 27 de Dic. de 2020
Comentada: Mohsin Shaikh el 28 de Dic. de 2020
x = 0:0.01:5; %% range of x values for graph
y = zeta_func(x);
plot(x,y),xlabel('x'), ylabel('zeta(x)'), title('Zeta(x) Graph'),
%% function definition
function zeta_val = zeta_func(x)
%% zeta(x) = summation of 1/x^n
%% here summing from n = 1 to 50
n = 1:1:50;
terms = 1./power(n,x); %% performing 1/x^i for all n
zeta_val = sum(terms); %% summing the values
end
I get the following error
Error using .^
Matrix dimensions must agree.
Error in zeta_gpu>zeta_func (line 11)
terms = 1./power(n,x); %% performing 1/x^i for all n
Error in zeta_gpu (line 2)
y = zeta_func(x);
Over here the function is being called on the whole array x. But I want to do element wise.
How do I acheive this over here?

Respuesta aceptada

Cris LaPierre
Cris LaPierre el 27 de Dic. de 2020
The error is because your vectors x and n are not the same length. x has 501 elements while n has 50. Also, power is doing n.^x. From how you worded things, perhaps you want x.^n?
Assuming you want every n to be raised to every x, I would look into using meshgrid. You can set it up so the rows represent unique values of x and the columns unique values of n (or vice versa). Then just sum by each x so you can create your plot.
x = 0:0.01:5; %% range of x values for graph
y = zeta_func(x);
plot(x,y),xlabel('x'), ylabel('zeta(x)'), title('Zeta(x) Graph'),
%% function definition
function zeta_val = zeta_func(x)
%% zeta(x) = summation of 1/x^n
%% here summing from n = 1 to 50
n = 1:1:50;
% create meshgrid, with unique x in each column, unique n in each row (50x501)
[X,N]=meshgrid(x,n);
terms = 1./power(N,X); %% performing 1/x^i for all n
zeta_val = sum(terms); %% summing the values
end
  3 comentarios
madhan ravi
madhan ravi el 27 de Dic. de 2020
Ah didn't see your comment Jan :), have to increase my font size ;)
Mohsin Shaikh
Mohsin Shaikh el 28 de Dic. de 2020
Thank you so much!

Iniciar sesión para comentar.

Más respuestas (2)

madhan ravi
madhan ravi el 27 de Dic. de 2020
terms = 1 ./ (n( : ) .^ x) % < 2016b 1 ./ bsxfun(@power, n( : ), x)

Luigi Emanuel di Grazia
Luigi Emanuel di Grazia el 27 de Dic. de 2020
Shouldn't it be power(x,n)?

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