Answer AlarmBell's Questions Thread (1 Viewer)

anomalousdecay

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Suppose polynomial p lies in and polynomial q lies in .

For a solid proof I would do the following:

Show for cases where n = 1 and m = 1, n = k, m = r.

Then use induction, n = k+1, m = r+1. State that if p and q are both continuous for all complex/real inputs (can't remember which course where you can assume all polynomials are in complex sets) then the composite is also continuous.

Then becomes a polynomial that lies in

As an engineer I would totally look at it and see what the resultant dimension for the manipulation is for any arbitrary m and n, then skip the proof part.

For b), check the definition for polynomials and mention continuity. You can use the result from part a) to help.

I could be wrong with a few things here as it's been a very long time since I touched this sort of maths.
 

AlarmBell

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Suppose polynomial p lies in and polynomial q lies in .

For a solid proof I would do the following:

Show for cases where n = 1 and m = 1, n = k, m = r.

Then use induction, n = k+1, m = r+1. State that if p and q are both continuous for all complex/real inputs (can't remember which course where you can assume all polynomials are in complex sets) then the composite is also continuous.

Then becomes a polynomial that lies in

As an engineer I would totally look at it and see what the resultant dimension for the manipulation is for any arbitrary m and n, then skip the proof part.

For b), check the definition for polynomials and mention continuity. You can use the result from part a) to help.

I could be wrong with a few things here as it's been a very long time since I touched this sort of maths.
Hey, thanks for your reply! I don't fully get it. Sorry but could you explain further with full working out please?
 

InteGrand

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Hey can anyone help me with these 2?

Both of these use the intermediate value theorem (IVT).

For 7), note that g(0) = f(0) – 0 = f(0), which is in [0, 1], since the range of f is in [0, 1]. If f(0) = 0, we have a suitable c of c = 0.

So suppose g(0) = f(0) is strictly positive instead (this is the only other possibility, since the range of f is [0, 1]). Now, g(1) = f(1) – 1. If f(1) = 1, we can use c = 1. Otherwise, we have f(1) < 1, since range(f) is a subset of [0, 1]. If f(1) < 1, then g(1) = f(1) – 1 < 0. So g(0) > 0 and g(1) < 0. Since g is continuous (being the difference of two continuous functions), it follows by the IVT that for some c in (0, 1), we have g(c) = 0, i.e. f(c) = c. So overall, we have f(c) = c for some c in [0, 1].

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For 8), since we are given that the limit of f(x) as x approaches infinity is -1, there must be an M > 0 such that f(x) is less than distance 1/2 (say) from -1 for all x > M (this follows from the definition of a limit to infinity). Hence for all x > M, we have f(x) < -1/2 < 0. So since f(0) = 1 > 0 and f(x) < 0 for some positive x > M, we have f(c) = 0 for some c in (0, inf), by the IVT (as f is given to be continuous). This proves the result.
 

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