/sci/ - Science & Math

>>8836021

xylnxy= xy(lnx+lny) = y*(xlnx)+x*(ylny)
D[xlnx]=1+ln(x)
1+ln(1)=1

QED

>>8836061
try and prove it without knowing f(x)=xlnx

f(xy) = yf(x) + xf(y)
Evaluate at x=1 and y=1:
f(1) = 2 f(1)
f(1) = 0

f(xy) = y f(x) + x f(y)
Take the derivate at y=1:
x f'(x) = f(x) + x
Solve the IVP:
f(x) = x ln(x)

>>8836083
i love you anon

>>8836021
I assume f is differentiable in around x =1.

First, differentiate the functional equation with respect to 'x', and we obtain: [math]yf'(xy) = yf'(x) + f(y), \ (2)[/math]
Set y = 1, we obtain: [math]f'(x) = f'(x) + f(1) [/math], but this implies that [math]f(1) = 0 [/math]. Next, substitute [math]y = 1/x [/math] into equation (2). We obtain: [math]\frac{1}{x} f'(1) = \frac{1}{x} f'(x) + f(\frac{1}{x}) [/math]. We know that f'(1) = 1, so multiply everything by 'x', to obtain: [math]1 = f'(x) + xf(\frac{1}{x}), \ (3) [/math]. Now, we go back to our original functional equation, and substitute in [math] y= \frac{1}{x}[/math]. We obtain: [math]f(1) = \frac{1}{x} f(x) + xf(\frac{1}{x}) [/math]. However we have already deduced that f(1) = 0. Therefore, [math]xf(\frac{1}{x}) = -\frac{1}{x}
f(x)[/math]. Therefore, we use this information for equation (3). We obtain: [math]1 = f'(x) - \frac{1}{x} f(x) [/math]

Now, this is a differential equation we can solve, we have that: [math] 1 - \frac{f}{x} = \frac{df}{dx} [/math]. This is a difficult differential equation to solve, so we cleverly (see: guided by our solution, or perhaps we see the 'x' in the denominator and we want to get rid of it), set [math]f(x) = x g(x) [/math], then [math]f'(x) = g(x) + xg'(x) [/math]. Putting this into our differential equation yields: [math] 1 + g(x) = g(x) + x g'(x) [/math], so: [math]g'(x) = \frac{1}{x} [/math], giving [math]g(x) = \ln x + c [/math].

So we have: [math]f(x) = x (\ln x + c) [/math], seeing that f(1) = 0, we have that c = 0. So: [math]f(x) = x \ln x [/math] QED

>>8836092
jeez man thanks a lot that helps

>>8836021
Let [math]g(x) = f(x)/x[/math]. g satisfies [math]g(xy) = g(x)+g(y)[/math] for each x,y > 0. If f is continuous, then g is of the form c*ln, with c a constant, and hence f is of the form f(x) = c*x*ln(x). Using the fact that f'(1)=1, we get f(x)=xln(x)

>>8836092
you didnt even need to assume, you are told f'(1) = 1

>>8836650
you have to prove that f(x) is deferentiable so yes you need it for that boii