Skip to main content
Bumped by Community user
Bumped by Community user
added 28 characters in body
Source Link
xskxzr
  • 7.6k
  • 5
  • 23
  • 47

Suppose we are give a natural number n$n$, the value of sin(x)$\sin(x)$ and cos(x)$\cos(x)$. How efficiently we can we compute sin(n x)
My$\sin(n x)$?

My Thoughts :
The sin (n x)$\sin (n x)$ expansion will have O(n)$O(n)$ terms the. The power terms will take log(n)$\log(n)$ time each to compute. But there will be a term nC_n/2$nC_n/2$ so if n 10$n=10$ this will be 10/5 How$10/5$. How to find the complexity of this term.? Is it Theta(2^n).$\Theta(2^n)$? Is there any alternate algorithm to compute it more efficiently? This way it looks around 2^n * logn * n$2^n n\log n $.

Suppose we are give a natural number n the value of sin(x) and cos(x). How efficiently we can compute sin(n x)
My Thoughts :
The sin (n x) expansion will have O(n) terms the power terms will take log(n) time each to compute. But there will be a term nC_n/2 so if n 10 this will be 10/5 How to find the complexity of this term. Is it Theta(2^n). Is there any alternate algorithm to compute it more efficiently? This way it looks around 2^n * logn * n

Suppose we are give a natural number $n$, the value of $\sin(x)$ and $\cos(x)$. How efficiently can we compute $\sin(n x)$?

My Thoughts :
The $\sin (n x)$ expansion will have $O(n)$ terms. The power terms will take $\log(n)$ time each to compute. But there will be a term $nC_n/2$ so if $n=10$ this will be $10/5$. How to find the complexity of this term? Is it $\Theta(2^n)$? Is there any alternate algorithm to compute it more efficiently? This way it looks around $2^n n\log n $.

Source Link

Complexity of calculating sin(nx)

Suppose we are give a natural number n the value of sin(x) and cos(x). How efficiently we can compute sin(n x)
My Thoughts :
The sin (n x) expansion will have O(n) terms the power terms will take log(n) time each to compute. But there will be a term nC_n/2 so if n 10 this will be 10/5 How to find the complexity of this term. Is it Theta(2^n). Is there any alternate algorithm to compute it more efficiently? This way it looks around 2^n * logn * n