Answer AlarmBell's Questions Thread (1 Viewer)

AlarmBell · Mar 26, 2016

Hey there can anyone help me with this question? [emoji4]

Cheers,
AlarmBell.

anomalousdecay · Mar 26, 2016

Suppose polynomial p lies in $\mathbb{P}_n$ and polynomial q lies in $\mathbb{P} _m$ .

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 $p \circ q$ becomes a polynomial that lies in $\mathbb{P} _ {nm}$

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 · Mar 26, 2016

anomalousdecay said:
Suppose polynomial p lies in $\mathbb{P}_n$ and polynomial q lies in $\mathbb{P} _m$ .

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 $p \circ q$ becomes a polynomial that lies in $\mathbb{P} _ {nm}$

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 · Mar 26, 2016

AlarmBell said:
Hey, thanks for your reply! I don't fully get it. Sorry but could you explain further with full working out please?

$\noindent Here's the idea for showing that if $p$ and $q$ are polynomial functions, then $p\circ q$ is a polynomial function.$

$\noindent If either $p$ or $q$ is a constant polynomial, then so is $p\circ q$, so it is also a polynomial. Suppose that neither $p$ are constant polynomial functions. Then letting the degrees of $p$ and $q$ be $n$ and $m$ respectively, we can write $q(x) = \sum_{k=0} ^{m} a_{k} x^{k}$ and $p(y)=\sum_{j=0} ^{n} b_{j} y^{j}$. Hence$

$\begin{align*} \left(p \circ q \right) (x) &= p\left(q(x)\right) \\ &= \sum_{j=0} ^{n} b_{j} \left(q(x)\right)^{j} \\ &= \sum_{j=0} ^{n} b_{j} \left(\sum_{k=0} ^{m} a_{k} x^{k}\right)^{j}, \end{align*}$

$\noindent and if we expand everything out via the binomial theorem and group the terms, we clearly get a polynomial in $x$. Hence $p \circ q$ is a polynomial function.$

$\noindent As an example, if $q(x) = x^2 + 1$ and $p(y) = 4y^3 + 2y$, then $p\left(q(x)\right) = 4\left(x^2 + 1\right)^3 + 2\left(x^2 + 1\right)$, which is clearly a polynomial.$

$\noindent For the case where $p$ and $q$ are rational functions (which means they are each the ratio of two polynomials), $p \circ q$ may not even be defined anywhere. For instance, if $q(x)$ is the zero polynomial and $p(y) = \frac{1}{y}$, then $\left(p \circ q\right)(x)=\frac{1}{0}$, which is undefined always. However, it can be shown that $p\circ q$ is a rational function if it isn't something like the above example where it is never defined. The basic idea to show this could be to write $p(y)$ and $q(x)$ as ratios of two polynomial expressions, then $p\circ q$ (whose domain will be all those $x$ in the domain of $q$ such that $q(x)$ is in the domain of $p$) can also have its value at $x$ be written out in terms of sums etc. like for the previous question. Then on combining fractions and simplifying etc., everything will come out to be a ratio of two polynomials, so $p \circ q$ will be rational.$

AlarmBell · Mar 26, 2016

Thank you very much InteGrand!

AlarmBell · Mar 26, 2016

Hey can anyone help me with these 2?

InteGrand · Mar 27, 2016

AlarmBell said:
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].

----------------

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.

AlarmBell · Mar 27, 2016

Thank you InteGrand! <3 Much appreciated.

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