Tilted Forum Project Discussion Community - View Single Post - Intermediate Value Theorum

Yakk · 04-08-2004, 07:59 AM

As a general guild, the proof of a theorem often gives hints asto how it can be used. Try reading through the proof of the IVT in order to find an answer.

A solution using your version of the IVT would be something like:
Let f(x) = x + sin(x)
We are required to prove 1 is an element of f([0, π/6]). (f[0,π/6]) is the image under f of [0, π/6] See Footnote1 for a more fomal treatment of "image under").
Let A := f([0, π/6])

Let m be the minimium value of A. Let M be the maximium value of A. (note A = [m,M], by IVT(2) and IVT(3).)

Now, f(π/6) = 1.5. Thus, M >= 1.5.
f(0) = 0. This, m <= 0.

From this we know [0, 1.5] is a subset of [m, M].

Now, A = [m,M] (by the intermediate value theorem, part 2). 1 is an element of [0,1.5], which is a subset of A, which is f([0,π/6]). By the definition of f([0,π/6]), there exists an x in [0, π/6] such that f(x) = 1.

As an aside, we never discovered what m and M are: we only found an inequality restricting their values. We don't need to know what their value is in order to find them useful.

Footnotes:
1: If g is a function and B is a set, then "the image under g of B" is defined to be all the values {g(x) such that x is an element of B}. This is denoted g(B).

In other words, y is in g(B) if and only if there exists an x in B such that g(x) = y.

A property of images under functions is Monotonicity: if B is a subset of C, then g(B) is a subset of g(C). (not nessicarially a strict subset, but a subset). Ie, if y is an element of g(B), y is an element of g(C).

04-08-2004, 07:59 AM	#4 (permalink)
Yakk Wehret Den Anfängen! Location: Ontario, Canada	As a general guild, the proof of a theorem often gives hints asto how it can be used. Try reading through the proof of the IVT in order to find an answer. A solution using your version of the IVT would be something like: Let f(x) = x + sin(x) We are required to prove 1 is an element of f([0, π/6]). (f[0,π/6]) is the image under f of [0, π/6] See Footnote1 for a more fomal treatment of "image under"). Let A := f([0, π/6]) Let m be the minimium value of A. Let M be the maximium value of A. (note A = [m,M], by IVT(2) and IVT(3).) Now, f(π/6) = 1.5. Thus, M >= 1.5. f(0) = 0. This, m <= 0. From this we know [0, 1.5] is a subset of [m, M]. Now, A = [m,M] (by the intermediate value theorem, part 2). 1 is an element of [0,1.5], which is a subset of A, which is f([0,π/6]). By the definition of f([0,π/6]), there exists an x in [0, π/6] such that f(x) = 1. As an aside, we never discovered what m and M are: we only found an inequality restricting their values. We don't need to know what their value is in order to find them useful. Footnotes: 1: If g is a function and B is a set, then "the image under g of B" is defined to be all the values {g(x) such that x is an element of B}. This is denoted g(B). In other words, y is in g(B) if and only if there exists an x in B such that g(x) = y. A property of images under functions is Monotonicity: if B is a subset of C, then g(B) is a subset of g(C). (not nessicarially a strict subset, but a subset). Ie, if y is an element of g(B), y is an element of g(C). __________________ Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.