Problem with steps, limits and symbolic variable
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Hello. I'm trying to calculate this limit with matlab.
I wrote this function, and it works just fine if I do something like f(3)
f = @(x) (sum([1:2:2*x-1]))/(3*x^2);
However, trying to calculate limit yields an error "Unable to compute number of steps from 1 to 2*n - 1 by 2".
limit(f(x), x, 5)
I googled a little, and I think I understand why this happens, but I don't know how I can fix it.
Can someone help?
采纳的回答
Use symsum.
syms n
limit(symsum(2*n-1,0,n)/(3*n^2),inf)
As you can see, I used symsum to compute the sum. Then divide by (3*n^2), and take the limit, as n-->inf.
If you wanted to do this without the symbolic toolbox, well then you need to be able to take limits and infintie sums on double precision vectors. This really is something that is best done using symbolic tools. So without those tools, I would do exactly the same thing, but now the work must be done using pencil and paper, because you really do want to work with limits.
If I go back, and read some of your comments, I see your problem arises because you are trying to use the function limit on a colon operator. MATLAB cannot do that. Instead, you really need to compute that sum in some symbolic form, thus best done with symsum.
更多回答（1 个）
Walter Roberson
2020-12-12
0 个投票
Consider the sum term alone and find what it is for n=4. Now find it for n=5. And now for n=6. Do you see a pattern? Evaluate the full expression for those values; what do you observe? How many terms do you estimate it would take before any asymptope would be effectively negligible?
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