Solving A Functional Equation | Substitution

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[ManjulaMathew-wbzn]

In the second method, inside the parenthesis, the expression simplifies to 1 + cosec (theta) and lets equal it to t. So cosec^2theta = (t-1)^2 = 1+cot^2theta
cot^2theta =t^2-t taking the reciprocal of the square root, tan (theta) = 1/sqrt(t^2-t)
Substituting in the reduced original equation, f(t) = 1/sqrt(t^2-t)
Or f(x) = 1/sqrt(x^2-x)

[pageegap]

Expression of f(x) should consider 2 branches as mentioned by @yoav613 because original expression can lead to positives and negatives values.
Nice problem when considering substitution and fields of values.

[yoav]

f(x)=1/sqrt(x^2-2x) is for positive x, and f(x)=-1/sqrt(x^2-2x) is for negative x.

[GreenMeansGOF]

We have f(1+csc(θ))=tan(θ). Thus,
f(x)=tan(arccsc(x-1))
=sgn(x-1)/sqrt(x^2-2x).

[Ricardo_S]

If we have f(something)=x it means that something is equal to the inverse function f^-1(x)=(x+√(1+x²))/x now you would only have to calculate the inverse function of that function

[nicolascamargo]

Substitution t=1+csc(θ)
θ=arccsc(t-1)

and tan(arccsc(t-1)),



Llamando arccsc(t-1)=x

t-1=(hypotenuse)/(opposite)

opposite=(hypotenuse)/(t-1)

adyacent=s











f(t)=1/[√(t²-2t)]

f(x)=1/[√(x²-2x)]

[vaggelissmyrniotis]

Shouldn't -1/sqrt(t*(t-2)) be a solution too?

[alextang]

After see this sqrt(1+x^2), I will think this problem can be solved using triangular function. 😉😉😉😉😉😉

[maxvangulik]

I happened to notice that’s an inverse trig or hyperbolic function, because I’ve worked those out before. Actually, that’s e^ one of them, with a factored-out i somewhere if it’s trig. Let’s find out which one, but first factor an x out of the top and bottom:
1+sqrt(1/x^2+1)=f^-1(x)
Now I’ll check which function that is.

e^2y-2xe^y-1=0
e^y=x+sqrt(x^2+1) hah! Got it. 4 tries. I deleted tanh, cosh, and sinh.
(e^arsinh(x))/x=f^-1(x)
(e^arsinh(f(x)))/f(x)=x
e^arsinh(f(x))=xf(x)
arsinh(f(x))=ln(xf(x))

f^2(x)=xf^2(x)/2-1/2x
(x/2-1)f^2(x)-1/2x=0

Done :) now which method was that i haven’t watched the video yet

[rorydaulton]

Using your solution f(x) = 1/sqrt(x^2-2x) does not result in x on the right side, but rather abs(x). To get x, note that t and x have the same sign since sqrt(1+x^2) is positive and greater in absolute value than x. So modify the solution to f(x) = sgn(x)/ sqrt(x^2-2x), where sgn is the sign function, yielding -1 for negative parameters, 0 for zero, and 1 for positive parameters. That results in x on the right hand side of your equation, as desired. All this can be checked in a graphing program (I used Geogebra).

[gfu]

by the second method would be: f((sinx+1)/sinx)=tanx pero so, we have f(t)=1/sqrt[t(t-2)].