Basj - 1 year ago 75

Python Question

In math, Taylor series are important to get approximations of functions, with polynomials of small degree.

I wanted to see how such an approximation can be helpful, for example in order to speed up computations. Let's use the famous Taylor series:

`log(1+x) = x + 0.5 * x^2 + (error term)`

Morally, computing the value of a polynomial of degree 2 should be much faster than computing a

`log`

Thus a code to test this:

`import numpy, time`

def f(t):

return t + 0.5 * t ** 2

f = numpy.vectorize(f)

s = time.time()

for i in range(100):

x = numpy.random.rand(100000)

numpy.log(1 + x)

print time.time() - s # 0.556999921799 seconds

s = time.time()

for i in range(100):

x = numpy.random.rand(100000)

f(x)

print time.time() - s # arghh! 4.81500005722 seconds

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Answer Source

With Python+Numpy, it's probably optimized here and there, and so it's impossible to really benchmark `log(1+x)`

vs `x + 0.5 * x^2`

.
So I moved to C++.

Time per operation with log: 19.57 ns

Time per operation with order-2 Taylor expansion of log: 3.73 ns

So roughly a x5 factor !

```
#include <iostream>
#include <math.h>
#include <time.h>
#define N (1000*1000*100)
#define NANO (1000*1000*1000)
int main()
{
float *x = (float*) malloc(N * sizeof(float));
float y;
float elapsed1, elapsed2;
clock_t begin, end;
int i;
for (i = 0; i < N; i++)
x[i] = (float) (rand() + 1) / (float)(RAND_MAX);
begin = clock();
for (i = 0; i < N; i++)
y = logf(x[i]);
end = clock();
elapsed1 = float(end - begin) / CLOCKS_PER_SEC / N * NANO;
begin = clock();
for (i = 0; i < N; i++)
y = x[i] + 0.5 * x[i] * x[i];
end = clock();
elapsed2 = float(end - begin) / CLOCKS_PER_SEC / N * NANO;
std::cout << "Time per operation with log: " << elapsed1 << " ns\n";
std::cout << "Time per operation with order-2 Taylor epansion: " << elapsed2 << " ns";
free(x);
}
```

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