Problem with steps, limits and symbolic variable

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?

 Réponse acceptée

John D'Errico
John D'Errico le 13 Déc 2020
Modifié(e) : John D'Errico le 13 Déc 2020
Use symsum.
syms n
limit(symsum(2*n-1,0,n)/(3*n^2),inf)
ans = 
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.

Plus de réponses (1)

Walter Roberson
Walter Roberson le 12 Déc 2020

0 votes

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?

1 commentaire

Egor Svito
Egor Svito le 13 Déc 2020
I'm well aware that this sum is basically n^2. But the question isn't about solving this particular limit, but rather about the problem with exactly the sum - that I don't know how to do it via matlab.

Connectez-vous pour commenter.

Catégories

En savoir plus sur Mathematics dans Centre d'aide et File Exchange

Produits

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by