arrayfun with different dimensions

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Dimitri
Dimitri 2014-5-4
评论： Jos (10584) 2014-5-7
Hi guys, I have the following question.
I am trying to calculate sum_(j=1)^k (sum_(i=1)^(k-1) ( (i+j)! ))
I defined a function f=@(k) sum(factorial(1:k)+factorial(1:k-1))
Then x=1:5 in order to evaluate the sum for the first 5 natural numbers
But arrayfun(f,x) produces a mistake since it says that I have different dimensions. Which is correct.. but why should it be a problem? And how should I evaluate the sum then?
Many thanks,Dimitri
  1 个评论
Star Strider
Star Strider 2014-5-4
It would help if you posted all your relevant code. Be sure to include your arrayfun call.

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采纳的回答

Jos (10584)
Jos (10584) 2014-5-4
factorial(1:k) will give you a vector with k values while factorial(1:k-1) will a vector with k-1 values. You simply cannot add vectors of unequal lengths ...

更多回答（4 个）

Jan
Jan 2014-5-4
Start with two simple for loops:
S = 0;
for jj = 1:k
for ii = 1:k-1
S = S + ...
end
end
Dimitri
Dimitri 2014-5-4
Guys, thanks a lot for all your comments. I understand the problem of unequal length, and so was looking for an alternative formulation. Jan thank you, I'll try it tomorrow. If it would not work I put the code, star strider.
Dimitri
Dimitri 2014-5-5
编辑：Dimitri 2014-5-5
Hi again. I decided to put the code to make it more clear. I have a simple case of a double sum where the first sum is running from 1..k, the second from 1..l, where k,l might have different sizes. The sum refers to factorial of (i+j):
sum_(i=1)^k sum_(j=1)^l (i+j)!
Ok, so here I define the sum:
f=@(k,l) sum( factorial((1:k)+(1:l)) )
x=1:5
y=1:4 %say, I want to evaluate the sum where the first index runs utp 5 and the second up to 4 Then:
arrayfun(f,x,y)
What am I doing wrong? Maybe I have to define differently the matrix (x,y) which has to be assigned to the sum?
Many thanks!
Jos (10584)
Jos (10584) 2014-5-6
So you want 20 sums in total (5 values of x combined with 4 values of y)?
x = 1:5
y = 1:4
[XX,YY] = ndgrid(x,y)
f = @(k) sum(factorial((1:XX(k))+(1:YY(k))) )
S = arrayfun(f,1:numel(XX))
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Dimitri
Dimitri 2014-5-7
Hi Jos, unfortunately it still shows the same mistake as before: that the "matrix dimensions must agree".
Jos (10584)
Jos (10584) 2014-5-7
Oops. Yes, of course, because again you want to add two vectors that are not equally long. This means that your formula is not right! Can you split your formula into sub-formulaes like this, and show the expected output for each step when you specify a specific a and b?
a = …
b = …
c = a + b % !! this errors when a and b do not have the same number of elements
d = factorial(c)
e = sum(d)

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