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Is there a formula?

From last time...

A mathematical question: Is there a simple formula for ∑0<m<nsin⁡(m)\sum_{0< m< n} \sin(m) similar to the formula ∑0<m<nm=n(n−1)2\sum_{0< m< n} m = \frac{n(n-1)}{2}?

YES!

Better yet, you can use Sage to discover it!

k, n = var('k,n')
show(sum(k, k, 1, n-1).factor()) ## Warning: in this notation, the right endpoint is *included*, contrary to usual Python syntax
12 (n−1)n\displaystyle \frac{1}{2} \, {\left(n - 1\right)} n
k, n = var('k,n')
show(sum(k^4, k, 1, n).factor())
130 (3 n2+3 n−1)(2 n+1)(n+1)n\displaystyle \frac{1}{30} \, {\left(3 \, n^{2} + 3 \, n - 1\right)} {\left(2 \, n + 1\right)} {\left(n + 1\right)} n
k, n = var('k,n')
f = sum(sin(k), k, 1, n-1)
show(f)
cos⁡(narctan⁡(sin⁡(1)cos⁡(1)))sin⁡(1)−(cos⁡(1)−1)sin⁡(narctan⁡(sin⁡(1)cos⁡(1)))−sin⁡(1)2 (cos⁡(1)−1)\displaystyle \frac{\cos\left(n \arctan\left(\frac{\sin\left(1\right)}{\cos\left(1\right)}\right)\right) \sin\left(1\right) - {\left(\cos\left(1\right) - 1\right)} \sin\left(n \arctan\left(\frac{\sin\left(1\right)}{\cos\left(1\right)}\right)\right) - \sin\left(1\right)}{2 \, {\left(\cos\left(1\right) - 1\right)}}

Aside: a couple of ways you could have discovered this formula yourself:

  • Use the identity sin⁡(x)+sin⁡(y)=⋯\sin(x) + \sin(y) = \cdots

  • Write sin⁡(x)\sin(x) in terms of eixe^{ix} and sum the resulting geometric series.



%timeit float(f(n=10000))
625 loops, best of 3: 663 µs per loop 
%timeit float(f(n=100000))
625 loops, best of 3: 341 µs per loop 
fast_float?

File: /projects/sage/sage-7.5/src/sage/ext/fast_eval.pyx
Signature : fast_float(f, old=None, expect_one_var=False, *args)
Docstring :
Tries to create a function that evaluates f quickly using floating-
point numbers, if possible.  There are two implementations of
fast_float in Sage; by default we use the newer, which is slightly
faster on most tests.

On failure, returns the input unchanged.

INPUT:

* "f"    -- an expression

* "vars" -- the names of the arguments

* "old"  -- use the original algorithm for fast_float

* "expect_one_var" -- don't give deprecation warning if "vars" is
  omitted, as long as expression has only one var

EXAMPLES:

   sage: from sage.ext.fast_eval import fast_float
   sage: x,y = var('x,y')
   sage: f = fast_float(sqrt(x^2+y^2), 'x', 'y')
   sage: f(3,4)
   5.0

Specifying the argument names is essential, as fast_float objects
only distinguish between arguments by order.

   sage: f = fast_float(x-y, 'x','y')
   sage: f(1,2)
   -1.0
   sage: f = fast_float(x-y, 'y','x')
   sage: f(1,2)
   1.0

g = fast_float(f, 'n')
g
<sage.ext.interpreters.wrapper_rdf.Wrapper_rdf object at 0x7fa97c11a788> 
%timeit g(10000)
625 loops, best of 3: 587 ns per loop 
g(10000)
1.9395054106807146 
N(f(10000))
1.93950541068071 
N((663 * 10^(-6)) / (587 * 10^(-9)))
1129.47189097104

1000x speedup!

%time float(sum(sin(j) for j in [1..10000]))
1.6338910217924534 CPU time: 2.08 s, Wall time: 2.12 s 
2.08 / (587*10^(-9))
3.54344122657581e6

So the net speedup from our first attempt is a factor of several million in speed.

Exercise now: Draw a plot of the function f(n)=∑0<m<nsin⁡(m)f(n) = \sum_{0< m< n} \sin(m) for various values of nn...

# one way
stats.TimeSeries([sin(n) for n in range(10000)]).sums().plot()

# Another -- use the formula above...



show(points([(n, g(n)) for n in range(10000)]))