2
$\begingroup$

I am trying to use Mathematica, in this case in a well known and rather simple list of equations, but I get stuck whit Mathematica. Since I am just starting to use Mathematica for mathematical purposes, I am aware that this might be to simple. However, I would like to get a good demonstration of how to proceed.

I am trying to follow: This page Normal Distribution

Here is what I have done so far.

D[f[x], x]

D[f, f] == -k (x - μ) \[DifferentialD]x

D[f, f] == -k (x - μ) \[DifferentialD]x
∫\[DifferentialD]f/f == ∫(x - μ) \[DifferentialD]x

Log[f] == x^2/2 - x μ (
Log[f] == x^2/2 - x μ + Log[c]
Solve[Log[f] == x^2/2 - x μ + Log[c], f]

This is where I get stuck, and if you look at the page that I have linked, you see that this is not the answer given there. Could someone who is capable provide a step by step solution of how to derive this. I think it would be helpful for many people just starting to use Mathematica for this purpose. I also think that having an example to look at would make it easier to find your own solutions later on in other cases.

$\endgroup$
1
  • $\begingroup$ first of all, I think there's an error in your second line: should be "1/f [DifferentialD]f == -k (x - [Mu]) [DifferentialD]x" according to the page. $\endgroup$
    – Phab
    Apr 8 '14 at 10:01
2
$\begingroup$

Try this ... I changed your code a bit:

eqn1 = f'[x] == -k (x - μ) f[x]

eqn2 = 1/f ⅆf == -k (x - μ) ⅆx

∫1/
   f[x] ⅆf[
     x] == ∫-k (x - μ) ⅆx

Exp[Log[f[x]]] == Exp[k (-(x^2/2) + x μ)]
(* you'll see it's the same as *)
DSolve[eqn1, f[x], x]

Gives for me:

Out1: $f'(x)=-k f(x) (x-\mu )$

Out2: $\frac{df}{f}=-k (x-\mu ) dx$

Out3: $\log (f(x))=k \left(\mu x-\frac{x^2}{2}\right)$

Out4: $f(x)=e^{k \left(\mu x-\frac{x^2}{2}\right)}$

Out5: $\left\{\left\{f(x)\to c_1 e^{k \left(\mu x-\frac{x^2}{2}\right)}\right\}\right\}$

$\endgroup$
2
  • $\begingroup$ Why are you 1/f. I understand that Df with regards to f equals 1. Which is 1/1 but do not get it in this case. $\endgroup$
    – ALEXANDER
    Apr 8 '14 at 13:36
  • $\begingroup$ @ALEXANDER It's just like in the link you gave: Multiply the first equation by dx and divide it by f (or f[x]) ... lefthandside and righthandside and you'll get the second equation. It has nothing to do with Integration so far. But now, if you integrate both sides you'll have to integrate 1/f*df on the lhs thats [Integral]1/f[x] [DifferentialD]f[x] $\endgroup$
    – Phab
    Apr 8 '14 at 13:56
0
$\begingroup$

Here is a stepwise approach:

sol = First@DSolve[f'[x] == -k (x - μ) f[x], f[x], x];

yields:

{f[x] -> E^(k (-(x^2/2) + x μ)) C[1]}

In this case to get to the normal PDF:

c1 = First@
  Solve[Integrate[f[x] /. sol, {x, -Infinity, Infinity}, 
     GenerateConditions -> False] == 1, C[1]]

yields:

{C[1] -> (E^(-((k μ^2)/2)) Sqrt[k])/Sqrt[2 π]}

Note the constant is product of usual normalization constant and constant term of polynomial in exponent. To get to desired form:

exp = (f[x] /. sol) /. c1
FullSimplify[
 ReplacePart[
   exp, {1, 2} -> Factor[Collect[Together[exp[[1, 2]]], k]]] /. 
  k -> 1/σ^2, 
 Assumptions -> (Element[σ, Reals] && σ > 0)]

yields:

E^(-((x - μ)^2/(2 σ^2)))/(Sqrt[2 π] σ)

The first step substitutes solutions and constants. The second step involves factorizing the polynomial after addding fraction and taking out constant then subsitution of k for reciprocal of variance.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.