Integrate e^-(x^2) (1 Viewer)

[gurmies]

If you cannot evaluate integrals via integration there is a different method called numerical integration.
For those lost HSC kids who don't know what the fuck it is, Trapezoidal Rule and the Simpson's Rule is a form of numerical integration, the latter is quite accurate.
Of course the more terms of summation the better the accuracy.
Another reason why with advances in mathematics and computing we are developing more deadly machines like Metal Gear.


Also, there is a Taylor Polynomial approximation of order 20 and because of this high degree the approximation should be very good and at least CAN be integrated.

[maths]e^{-x^{2}} \approx 1-x^{2}+\frac{1}{2!}x^{4}-\frac{1}{3!}x^{6}+\frac{1}{4!}x^{8}-\frac{1}{5!}x^{10}+\frac{1}{6!}x^{12}-\frac{1}{7!}x^{14}+\frac{1}{8!}x^{16}-\frac{1}{9!}x^{18}+\frac{1}{10!}x^{20}[/maths]

Taylor polynomials were invented by the same guy who invented integration by parts, Brooke Taylor who actually stabbed Newton in the head and took the credit because he couldn't stand seeing Newton being in the spotlight of mathematics any longer.

How do you derive taylor polynomials for certain expessions?

 

[Drongoski]

If f(x) is a function with derivatives of all orders throughout some interval containing 'a' as an interior point, then the Taylor series generated at x = a is:





For the special case of x = 0 we have the Maclaurin series:

 

[Drongoski]

For the special case of a=0 we have the Maclaurin series:



Edit

corrected x=0 to a=0 thanks to tacogym ...

 

[gurmies]

Awesome, thanks Drongoski

 

[Drongoski]

Awesome, thanks Drongoski

You are welcome. I deserve no credit as I just lifted it off a book.

But Maclaurin series was in my Cambridge Overseas Higher School Certificate Pure Maths syllabus eons ago. Ha ha.

 

[tacogym27101990]

For the special case of x=0 we have the Maclaurin series:

a=0

 

[Drongoski]

a=0

you are right; my oversight.

 

[untouchablecuz]

one question about taylor series

lets say you wanted to approximate:

S [16->12] e^(-x^2) dx

would you let a = 14 (or something around this point)?

 

[Studentleader]

I notice nobody mentioned the incredibly beautiful solution to the problem of integrating this function between -infinity and plus infinity.
Since there is no latex here, I won't write it, but you can read a nice walkthrough here:
Gaussian Integral -- from Wolfram MathWorld

It's called the Gaussian integral.
Now when you look at that technique (look at equations (1) to (8) ), you might be put off by the use of a double integral, and the change to polar coordinates. I guess if you've never seen either of those things it might be a bit too much to swallow; but if you at least have seen polar coordinates you should appreciate the genius of the idea.

Also of great interest is the fact that this is the basic functional form of the so-called "normal distribution" (or "Gaussian distribution") or bell curve which underlies all of modern statistics. If you do science at Uni you'll be swamped with these things.

Thats what I thought of doing (after I put it in Mathematica) :rofl2:

 

[Studentleader]

But Maclaurin series was in my Cambridge Overseas Higher School Certificate Pure Maths syllabus eons ago. Ha ha.

It was in the WA Calculus syllabus for a while I think - we had a practical assignment on it.

 

[gurmies]

choosing an 'a' closer to the region of the function you wish to approximate will give a better approximation for less sums.

for example, if we wish to approximate sin(x) for small values of x, a = 0 is a good choice as it wouldnt require as many terms to sum for a good approximation. notice on the graph below how the approximate taylor polynomial can deviate significantly from sin(x) for points further away from a=0:



one obvious potential drawback from not centering your taylor series of the function around 0 however, is that computing f^(k)(a) may not be as simple or you could make each term harder to integrate.

for example, before we had:

e^(-x^2) = sum [(e^a)*(-x^2 - a)^k / k!]

integrating the RHS with respect to x is not easy when 'a' does not equal 0 as we have (-x^2 - a)^k to integrate rather than simply -x^(2k).

if i were to write a program to calculate S [16->12] e^(-x^2) dx, id take a = 0 and just sum more terms. if summing terms is too slow, there are also other, more advanced methods for finding a different series which converges faster to the same limit

oic

 

[rama_v]

Wow, this thread is still active four years after I started it. Such a silly question I asked then.

I wish they taught Taylor series (and many other things) in high school maths, it would have been really useful, and would have simplified limit proofs unbelievably.

E.g. Behaviour of Sin (x)/x as x-> 0:

Sin(x) /x = (x - x³/3! +x⁵/5! ... )/x = 1 - x²/3! + ...

= 1 - 0 + 0 -...

=1

 

[JDUBBS2344]

I think the easiest way to integrate x e^(-x^2) is by substitution.

First left u = -x^2 then du = -2x dx then you can rearrange du=-2x dx to become
-(1/2)du=dx then the integration of

xe^(-x^2)dx becomes

e^u(-1/2)du which you can take out the constant of (-1/2)

results in [(-1/2)e^x^2] + C since this is an indefinite integral