Answered: "Is limn→∞1n(cos⁡{πn}+cos⁡{2πn}+…+cos⁡{nπn}) a Riemann sum?"

Karlie Mays

Answered question

2022-04-13

lim_{n \to \infty} \frac{1}{n} (\cos {\frac{π}{n}} + \cos {\frac{2 π}{n}} + \dots + \cos {\frac{n π}{n}})

a Riemann sum?

xxkrnjangxxed9q

Beginner2022-04-14Added 14 answers

The Riemann sum is given by

S = \sum_{i = 1}^{n} f (y_{i}) (x_{i} - x_{i - 1})

where

x_{i - 1} \leq y_{i} \leq x_{i}

. If you choose

f (y_{i}) = \cos (y_{i}), y_{i} = \frac{i π}{n}

and

x_{i} = \frac{i π}{n}

you get

\int_{0}^{π} \cos (x) d x = lim_{n \to {\infty}} \sum_{i = 1}^{n} \cos (\frac{i π}{n}) (x_{i} - x_{i - 1})

which is just the sum you have with an extra factor

π

. Therefore

\frac{1}{π} \int_{0}^{π} \cos (x) d x = lim_{n \to \infty} \frac{1}{n} (\cos {\frac{π}{n}} + \cos {\frac{2 π}{n}} + \dots \cos {\frac{n π}{n}})

Jameson Jensen

Beginner2022-04-15Added 16 answers

Assume that

x_{k - 1} \leq c_{k} \leq x_{k}, x_{0} = 0, x_{n} = π

, and that the interval of integration is divided into n sub-intervals of equal width. Under these circumstances

x_{k} - x_{k - 1} = \frac{π}{n}

and

lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} \cos \frac{k π}{n} = lim_{n \to \infty} \sum_{k = 1}^{n} \cos (c_{k}) \frac{x_{k} - x_{k - 1}}{π}

= lim_{n \to \infty} \sum_{k = 1}^{n} (\frac{1}{π} \cos (c_{k})) (x_{k} - x_{k - 1})

= \int_{0}^{π} \frac{1}{π} \cos x d x

The sum

\sum_{k = 1}^{n} (\frac{1}{π} \cos (c_{k})) (x_{k} - x_{k - 1})

is a Riemann Sum of the function

f (x) = \frac{1}{π} \cos x

in the interval

[0, π]

, but not of the function

\cos x

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