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This can be written as follows: A = sqrt(7 - sqrt(7 + A)) So, A^2 = 7 - sqrt(7+A) sqrt(7+A) = 7 - A^2 7 + A = 49 - 14*A^2 + A^4 A^4 - 14*A^2 - A + 42 = 0 (A-2)*(A+3)*(A^2 - A - 7) = 0 The quadratic factor has complex roots, so let's just neglect those. That leaves A=2 and A=-3 as solutions. Both of these work, but if you try to use A=-3 then when checking it you have to take the negative square root at one stage. So A=2 seems like the most "pleasing and natural" solution.
Far too complicated! Here's how to do it without a quartic. Let B be the analogous expression with + and - exchanged. Then A = sqrt(7-B) and B=sqrt(7+A). Square both and subtract: A^2-B^2+A+B=0. This means (A+B)(A-B+1)=0. As A>0 and B>0, one has B=A+1. Insert into A^2=7-B, so A^2+A=6, thus A=2 and B=3.
Let f(x)=sqrt(7-sqrt(7+x)). Then A has to satisfy the fixed point equation f(A)=A. So, let's now solve the equation sqrt(7-sqrt(7+x))=x. Just by trying a couple of values, we can easily see that x=2 is a solution. Are there other solutions? First of all, x has to be non-negative because it's equal to a square root. The equation can be written in the equivalent form 7-sqrt(7+x)=x^2. Now the two functions on the left-hand side of this latter equation and on the right-hand side have graphs which are arcs of parabolas in the plane. By an obvious geometric argument, these two graphs can intersect in at most one point in the x>0 half-plane. Therefore x=2 is the only solution to the fixed point equation, and A=2.
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to summarise, you just set the infinite bit to some variable so you don't have to think about it, then solve a quartic through factorisation
This can be written as follows:
A = sqrt(7 - sqrt(7 + A))
So,
A^2 = 7 - sqrt(7+A)
sqrt(7+A) = 7 - A^2
7 + A = 49 - 14*A^2 + A^4
A^4 - 14*A^2 - A + 42 = 0
(A-2)*(A+3)*(A^2 - A - 7) = 0
The quadratic factor has complex roots, so let's just neglect those. That leaves A=2 and A=-3 as solutions.
Both of these work, but if you try to use A=-3 then when checking it you have to take the negative square root at one stage. So A=2 seems like the most "pleasing and natural" solution.
Far too complicated! Here's how to do it without a quartic.
Let B be the analogous expression with + and - exchanged. Then A = sqrt(7-B) and B=sqrt(7+A). Square both and subtract: A^2-B^2+A+B=0. This means (A+B)(A-B+1)=0. As A>0 and B>0, one has B=A+1. Insert into A^2=7-B, so A^2+A=6, thus A=2 and B=3.
1:30 it's the range, not the domain, m is the output.
Let f(x)=sqrt(7-sqrt(7+x)). Then A has to satisfy the fixed point equation f(A)=A.
So, let's now solve the equation sqrt(7-sqrt(7+x))=x.
Just by trying a couple of values, we can easily see that x=2 is a solution. Are there other solutions? First of all, x has to be non-negative because it's equal to a square root. The equation can be written in the equivalent form 7-sqrt(7+x)=x^2. Now the two functions on the left-hand side of this latter equation and on the right-hand side have graphs which are arcs of parabolas in the plane. By an obvious geometric argument, these two graphs can intersect in at most one point in the x>0 half-plane. Therefore x=2 is the only solution to the fixed point equation, and A=2.