Think about the expansion, it is not small

Calculus Level 4

$\large \displaystyle L=\lim_{n \to \infty} \sum_{k=1}^{n} \dfrac{n-k}{n^2} \cos\left(\dfrac{4k}{n}\right)$

Find the value of $\lfloor 1000L \rfloor$ .

The answer is 103.

1 solution

Vishwak Srinivasan
Jul 7, 2015

The limit/summation can be converted into a Riemann Integral.

$\displaystyle \lim_{n\to \infty} \dfrac{1}{n} \sum_{k=1}^{n} \dfrac{n-k}{n}\cos\left(\dfrac{4k}{n}\right) = L$

$L = \displaystyle \int_{0}^{1} (1-x)\cos 4x dx = \dfrac{4(1-x)\sin 4x - \cos 4x}{16} \huge|_{0}^{1}$

$L = \dfrac{1-\cos 4}{16} \approx 0.1033 \Rightarrow \lfloor 1000L \rfloor = \boxed{103}$

Don't you think the answer would be 10 ? If L = 0.1033 , then 100L = 10.33, so floor(100L) = 10. I wasted my 3 attempts because of your wrong answer :/

Satyajit Mohanty - 5 years, 11 months ago

Thanks. Those who answered 10 have been marked correct.

In future, if you spot any errors with a problem, you can “report” it by selecting "report problem" in the “dot dot dot” menu in the lower right corner. This will notify the problem creator who can fix the issues.

Calvin Lin Staff - 5 years, 11 months ago

Definitely, sir.

Satyajit Mohanty - 5 years, 11 months ago

I am extremely sorry for the inconvenience. I will make amends right away. @Calvin Lin sir, can you mark those who have marked 10 initially as right.

Vishwak Srinivasan - 5 years, 11 months ago

Think about the expansion, it is not small

The answer is 103.

1 solution

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