Hello, i would need help for understanding something in the Riemann sums subject
#Riemann sums
noble lotus
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Hello, i would need help for understanding something in the Riemann sums subject
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noble lotus
i have to calculate the limit of this sequence using Riemann sum
But i don't know what to do... do i want to have a 1/n and k/n as usual ? does the fact that K goes from 1 to n-1 matters a lot ?
what are the terminals of the integral ?
this is what i've got
elder blade
this is what i've got
The approach is correct, but you forgot to write the limit.
noble lotus
Ah the limit of the sum is equal to the integral ?
elder blade
Yes.
noble lotus
But the result is good ? i have no idea what does the n-1 means
elder blade
But the result is good ? i have no idea what does the n-1 means
The upper limit of the sum.
We are summing the areas of n rectangles, after all.
noble lotus
no i know, but in the integral, does it changes anything ?
from 0 to 1
and why is it always from 0 to 1, since in the exercise nothing is said about the terminals
elder blade
No.
The left Riemann sum for ∫(f(x)dx, a, b) with n equal intervals is ((b - a)/n)Σ(f(a + k(b - a)/n), k = 0 to n - 1).
In our case a = 0, b = 1, so the sum becomes (1/n)Σ(f(k/n), k = 0 to n - 1). Note that we skip the term for k = 0 since f(0) = √(0) = 0.
elder blade
It's easier if you try drawing what it looks like.
Note that the right sum is the same, but k changes from 1 to n.
noble lotus
elder blade
Yes.
Now, make a rectangle on each interval, its height being the value of the function on the left side of the interval.
noble lotus
elder blade
Yes.
The sum of areas of those rectangles is the left Riemann sum for ∫(f(x)dx, 0, 1).
elder blade
Well, it could also be the right Riemann sum.
And, while I haven't seen exercises utilizing it, we aren't required to take intervals of equal length in a Riemann sum.
noble lotus
i got this, and yesterday my teacher told me about the 1/n not being correct since the last term is 2n
so in this case it would still be between 0 and 1, but we want a 1/2n ?
elder blade
Well, the integral will be from 0 to 2, then. Just be careful with the constant.
elder blade
We have:
k/(n^2 + k^2) = (1/n)((k/n)/(1 + (k/n)^2)) = 2(1/(2n))((k/n)/(1 + (k/n)^2))
So, this is a Riemann sum for ∫(f(x)dx, 0, 2), where f(x) = 2x/(1 + x^2).