What is the weight of a square on this hanger if a triangle weighs 3 grams?
1 answer:
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We have to find the values of F.
In this case. F is unlikely to be a polynomial.
But the problem is, we can’t calculate the values of F directly.
There is no real value of x for which x = x−1 x because F isn’t defined at 0 or 1. so,
substituting x = 2.
F(2) + F(1/2) = 3.
Substitute, x = 1/2
F(1/2) + F(−1) = −1/2.
We still are not getting the required value,
therefore,
Substitute x = −1
As, F(2) +F(−1) = 0.
now we have three equations in three unknowns, which we can solve.
It turns out that:
F(2) = 3/4
F(3) = 17/12
F(4) = 47/24
and
F(5) = 99/40
Setting
g(x) = 1 − 1/x
and using
2 → 1/2
to denote
g(2) = 1/2
we see that :
x → 1 - 1/x → 1/(1-x) →x
so that:
g(g(g(x))) = x.
Therefore, whatever x 6= 0, 1 we start with, we will always get three equations in the three “unknowns” F(x), F(g(x)) and F(g(g(x))).
Now solve these equations to get a formula for F(x)
As,
h(x) = (1+x)/(1−x)
which satisfies h(h(h(h(x)))) = x
Now, mapping x to h(x) corresponds to rotating the circle by ninety degrees.
In this case. F is unlikely to be a polynomial.
But the problem is, we can’t calculate the values of F directly.
There is no real value of x for which x = x−1 x because F isn’t defined at 0 or 1. so,
substituting x = 2.
F(2) + F(1/2) = 3.
Substitute, x = 1/2
F(1/2) + F(−1) = −1/2.
We still are not getting the required value,
therefore,
Substitute x = −1
As, F(2) +F(−1) = 0.
now we have three equations in three unknowns, which we can solve.
It turns out that:
F(2) = 3/4
F(3) = 17/12
F(4) = 47/24
and
F(5) = 99/40
Setting
g(x) = 1 − 1/x
and using
2 → 1/2
to denote
g(2) = 1/2
we see that :
x → 1 - 1/x → 1/(1-x) →x
so that:
g(g(g(x))) = x.
Therefore, whatever x 6= 0, 1 we start with, we will always get three equations in the three “unknowns” F(x), F(g(x)) and F(g(g(x))).
Now solve these equations to get a formula for F(x)
As,
h(x) = (1+x)/(1−x)
which satisfies h(h(h(h(x)))) = x
Now, mapping x to h(x) corresponds to rotating the circle by ninety degrees.