How do you solve this problem analytically?
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How do you solve this problem analytically?
here's a less shitty pic
>>8002768
start with substitution y = x+1
x^y = y^x
ln(x^y) = ln(y^x)
y*ln(x) = x*ln(y)
y/x = ln(y)/ln(x)
Now, you cry.
exp((x+1)ln(x))=exp(xln(x+1))
=>(x+1)ln(x)=xln(x+1)
it's easy after that
logs
>>8002768
hard
x^(x+2) = (x+2)^x has a very simple solution though:
x=2
>>8002780
maybe I'm just brain-dead today, but i don't see what you would do next.
well i'll end it for you
assuming that x=/=0 because 0 is not a solution,
=>ln((x^(x+1))/(x+1)^x)=0
=>ln(x(x/x+1)^x))=0
=>ln(x) + x(ln(x/x+1) = 0
study a fonction to end it.
>>8002768
x=(1+1/x)^x
x~=e
https://www.wolframalpha.com/input/?i=solve+x%3D(1%2B1%2Fx)%5Ex
>x=2.29317~
>>8002783
prove that's the only solution tho
>>8002907
complex solution/s could exist but im no math
for find real one yourself
make a implicit term, decide for a numerical operation(newton, secant,...), if derivative is necessary decide for analytical/numerical derivative, iterate shit
or youre using excel solver for finding x with criteria archieving implicit term (...=0) as pic related
Mathematica's FindInstance function isn't showing anything, so presumably you can't get a closed form representation of the solution without advanced techniques.
>>8002946
sry read over analytical or rather forget about while reading through.
im curious about if there is a complex solution to this
>>8002992
I don't know man. Complex exponents are wierd. Anyone ever show you this one? [eqn]i^i = e^{-\frac{\pi}{2}}[/eqn]
>>8003030
read that some weeks ago on /sci/.
you can transform it simply into
-i ln(-1) = Pi
this is a bit more tolerable ;)
man, im happy that im no math or EE.
>>8002907
-4 is also a solution
>>8002907
study the function ln(x)/x
this is highschool bro.
>>8003030
My personal favorite is [math]\sqrt{j} = \frac{1}{\sqrt2} + j\frac{1}{\sqrt2}[/math]
>>8003030
There is no negative in e's exponent.
Remove the negative in i's exponent and your equality will be true.
>>8003245
>using j as imaginary unit
Spot the EE.
>>8003245
That's like saying "2+2=5, but the 5 is just a different symbol for 4!"
i has an assigned value. If you change the symbol, the value changes and you're just talking nonsense.
>>8003268
You don't need to introduce the complex log to do that calculation. Also ln is undefined for imaginary numbers.
i = e^(i pi/2) so i^i = e(-pi/2). This is basic complex analysis and would probably impress freshmen and engineers.
>>8003245
Ahh, the jmagjnary number
>>8002946
>german engineering
[math] x^{x+1} = (x+1)^x [/math]
[math] x = \left(1 + \frac{1}{x} \right)^x [/math]
Consider the function [math]f: [2,e] \to [2,e] [/math] with [math]f(x) = \left(1 + \frac{1}{x} \right)^x [/math].
[math] |f'(x)| < 1 [/math] so by Banach's fixed point theorem the sequence [math](x_k) [/math] with [math]x_0 = 2 [/math] and [math] x_{k+1} = f(x_k) [/math] converges against a solution of [math] x^{x+1} = (x+1)^x [/math].
>>8002868
>x=2.29317~
>~=e
wat
[eqn]x^{y} = y^{x}[/eqn]
[eqn]x^{y} - y^{x} = 0[/eqn]
where y = x + 1
That's all you need to do anything with it. Next!
>>8003299
>Also ln is undefined for imaginary numbers
You're pretty stupid.
>>8002768
x+1=x^(1+1/x)
x=x^(1+1/x)-1
x=(x^(1+1/x)-1)^(1+1/x)-1
Uhm...
