please share matlab command please
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3 comentarios
parid
el 7 de En. de 2023
Atsushi Ueno
el 7 de En. de 2023
Editada: Atsushi Ueno
el 7 de En. de 2023
syms n x
y = symsum ((1+n^2/n),n,0,100)
syms n x
y = symsum ((1+n^2)/n,n,0,100)
You can avoid the division by 0 error.
syms n
f = (1+n^2)/n
%break the sum into two parts
part2 = symsum(f, n, 1, 100)
part1 = limit(f, n, 0)
y = part1 + part2
Respuestas (1)
Walter Roberson
el 7 de En. de 2023
1 voto
You can break this up into two sums: one for n = 0 exactly, and the other one for n = 1 to 100.
You can calculate the 1 to 100 using symsum
The value for 0 exactly would normally involve a division by 0, but you can try to approach the problem using limit . Just make sure to compare the limit from the "left" with the limit from the "right"
4 comentarios
parid
el 7 de En. de 2023
Walter Roberson
el 7 de En. de 2023
You coded to calculate n^2 and divide by n, and add 1 to the result of the division. But the expression requires you to take n^2 and add 1 to the result, and divide the total by n
Mathematically except at 0 the expression is the same as (1/n)+n
parid
el 7 de En. de 2023
Remember that / has higher priority than division, so
1+n^2/n is 1 + (n^2/n)
In turn, ^ has higher priority than / so this is 1 + ((n^2)/n)
But n^2/n is just n unless n is 0 (or unless n is so large that n^2 overflows to infinitity), so that expression is equivalent to 1 + (n)
This is different than taking n^2 and adding 1 to that value, and then dividing the whole thing by n .
Consider for example if n = 2 and n = 5
n = [2 5]
result1 = (1+n.^2./n)
result2 = 1 + ((n.^2)./n)
temporary = n.^2 + 1; result3 = temporary ./ n
result1 and result2 are the same -- they are the same calculation. result3 is different: they show what happens when you make sure to do the addition first before the division.
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