MATH1251 Questions HELP (5 Viewers)

InteGrand · Sep 15, 2016

$\noindent \textbf{Remarks.} Note that this Vandermonde Matrix is exactly the coefficient matrix we'd get if we were trying to fit an (at-most) $(n-1)$ degree polynomial $a_{0} + a_1 t + \dots + a_{n-1}t^{n-1}$ to the points $\left(t_{1},y_{1}\right),\ldots , \left(t_{n},y_{n}\right)$, where the $y_{i}$ are given $y$-values for the polynomial (and we are trying to find coefficients $a_0,\ldots , a_{n-1}$ so that the polynomial fits these points exactly).$

$\noindent The above product formula shows that as long as our data points are pairwise distinct $t$-values, we are guaranteed to be able to fit an (at-most) $(n-1)$ degree polynomial to these points (of course, if the points \emph{don't} have pairwise distinct $t$-values, i.e. if there is a pair with the same $t$-value, then we can't fit any function at all to the points, since we'd have a $t$-value with two different $y$-values). This is because pairwise distinct $t$ values implies the product is non-zero, so the determinant is non-zero, so there is a solution for the polynomial coefficients $a_i$ (and is also unique).$

$\noindent A probably easier way to just show there is a solution is to just construct one. You can search up the Lagrange Interpolating Polynomial for this:$

https://en.wikipedia.org/wiki/Lagrange_polynomial .

$\noindent Of course, uniqueness is easier to show. If such a polynomial exists, it is unique because as we know a polynomial of degree at-most $(n-1)$ is uniquely determined by $n$ points (if there were two such polynomials, their difference, an at-most $n-1$ degree polynomial, would be $0$ at these $n$ data points, and we know that a polynomial of degree at-most $n-1$ with $n$ roots must be the zero polynomial, as can be shown using the factor theorem and induction for example.)$

leehuan · Sep 15, 2016

Oh right yeah. This question did come out of the lot (and by 'lot' I really just mean three questions) under vector spaces in the polynomial interpolation section.

1008 · Sep 18, 2016

Hey everyone, I've got another question:

$\\\text{Solve }x^2y''-2xy'+2y=0\text{ by:}\\\text{a) letting }x=e^t\\\text{b) letting }y=x^m$

leehuan · Sep 18, 2016

1008 said:
Hey everyone, I've got another question:

$\\\text{Solve }x^2y''-2xy'+2y=0\text{ by:}\\\text{a) letting }x=e^t\\\text{b) letting }y=x^m$

Didn't your tutor go over that one with you?

$x=e^t \implies \frac{dx}{dt}=e^t$

$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=e^{-t}\frac{dy}{dt}$

$\frac{d^2y}{dx^2}=\frac{d}{dx} \left( \frac{dy}{dx} \right) =\frac{d}{dx}\left( e^{-t}\frac{dy}{dt} \right)$

$\stackrel{implicitdiff}{\implies}\\ \frac{d^2y}{dx^2}=e^{-t}\frac{d}{dt}\left( e^{-t}\frac{dy}{dt}\right)\\ \text{apply product rule and sub back in}$

And the other one just gives you a new characteristic equation in m

1008 · Sep 18, 2016

leehuan said:
Didn't your tutor go over that one with you?

$x=e^t \implies \frac{dx}{dt}=e^t$

$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=e^{-t}\frac{dy}{dt}$

$\frac{d^2y}{dx^2}=\frac{d}{dx} \left( \frac{dy}{dx} \right) =\frac{d}{dx}\left( e^{-t}\frac{dy}{dt} \right)$

$\stackrel{implicitdiff}{\implies}\\ \frac{d^2y}{dx^2}=e^{-t}\frac{d}{dt}\left( e^{-t}\frac{dy}{dt}\right)\\ \text{apply product rule and sub back in}$

And the other one just gives you a new characteristic equation in m

No he didn't

Wait, how'd you do the implicit diff part?

leehuan · Sep 18, 2016

1008 said:
No he didn't
Wait, how'd you do the implicit diff part?

d/dx ...
= d/dt ... * dt/dx

and dt/dx = exp(-t)

(Recall that implicit differentiation is just a fancy application of the chain rule)

1008 · Sep 18, 2016

leehuan said:
d/dx ...
= d/dt ... * dt/dx

and dt/dx = exp(-t)

(Recall that implicit differentiation is just a fancy application of the chain rule)

Thanks

leehuan · Sep 18, 2016

Paradoxica · Sep 18, 2016

leehuan said:
$\text{Let }\{a_n\},\{ b_n\}\text{ be sequences and suppose that }a_k\neq -b_k\\ \text{Suppose} \sum_{k=a}^\infty(a_k-b_k) \text{ converges}\\ \text{Must }\sum_{k=a}^\infty a_k\text{ converge?}$

No.

Example: a_n = 1/n, b_n = log(1+1/n)

sum (a_n - b_n) = γ (EulerGamma)

but sum (a_n) = ∞

leehuan · Sep 18, 2016

Paradoxica said:
No.

Example: a_n = 1/n, b_n = log(1+1/n)

sum (a_n - b_n) = γ (EulerGamma)

but sum (a_n) = ∞

Not that I know what EulerGamma is but good enough, thanks.
_______________________________________________

$\text{Show that this is }\textit{divergent...}\\ \sum_{k=2}^\infty \frac{(-1)^k}{\sqrt{k}+(-1)^k}$

They rearrange it into this and just say oh the latter sum is divergent so it must be divergent. Is this, on the other hand, acceptable?

$\sum_{k=2}^\infty \frac{(-1)^k\sqrt{k}}{k-1}-\sum_{k=2}^\infty \frac{1}{k-1}$

Paradoxica · Sep 18, 2016

I don't think that's valid reasoning. What you can do is show the former sum converges [it does converge, by the comparison test with Zeta(½)], and the latter diverges.

1008 · Sep 18, 2016

I had another question:

$\\\text{The limit of a recursively defined sequence.}\\\text{Suppose that }a_1=1 \text{ and } a_{n+1}=\sqrt{1+a_n}\text{ whenever }n\geq1\\\text{a) Show that }\sqrt{1+x}\in [1,2]\text{ whenever }x\in [1,2]\\\text{b) Use induction to show that the sequence }\{a_n\}\text{ is bounded.}\\\text{c) Use induction to show that }\{a_n\}\text{ is an increasing sequence.}\\\text{d) Explain why } \lim_{n\to \infty } a_n\text{ exists.}\\\text{e) Find }\lim_{n\to \infty } a_n.\text{ That is, find }\sqrt{1+\sqrt{1+\sqrt{1+ \cdot \cdot \cdot }}}$

I get (a), but nothing else...

InteGrand · Sep 18, 2016

1008 said:
I had another question:

$\\\text{The limit of a recursively defined sequence.}\\\text{Suppose that }a_1=1 \text{ and } a_{n+1}=\sqrt{1+a_n}\text{ whenever }n\geq1\\\text{a) Show that }\sqrt{1+x}\in [1,2]\text{ whenever }x\in [1,2]\\\text{b) Use induction to show that the sequence }\{a_n\}\text{ is bounded.}\\\text{c) Use induction to show that }\{a_n\}\text{ is an increasing sequence.}\\\text{d) Explain why } \lim_{n\to \infty } a_n\text{ exists.}\\\text{e) Find }\lim_{n\to \infty } a_n.\text{ That is, find }\sqrt{1+\sqrt{1+\sqrt{1+ \cdot \cdot \cdot }}}$

I get (a), but nothing else...

Here's some hints.

$\noindent b) It follows immediately from a) and induction that $1\leq a_n \leq 2$ for all $n$.$

$\noindent c) Inductive step: $a_{n+1} = \sqrt{1 + a_n} \geq \sqrt{1 + a_{n-1}} = a_n$. The inequality comes from the inductive hypothesis $a_n \geq a_{n-1}$ and since the function $\sqrt{1+x}$ is increasing.$

$\noindent d) Monotone Convergence Theorem.$

$\noindent e) The limit is the \textsl{Golden Ratio} $\varphi = \frac{1 + \sqrt{5}}{2}$.$

1008 · Sep 18, 2016

InteGrand said:
Here's some hints.

$\noindent b) It follows immediately from a) and induction that $1\leq a_n \leq 2$ for all $n$.$

$\noindent c) Inductive step: $a_{n+1} = \sqrt{1 + a_n} \geq \sqrt{1 + a_{n-1}} = a_n$. The inequality comes from the inductive hypothesis $a_n \geq a_{n-1}$ and since the function $\sqrt{1+x}$ is increasing.$

$\noindent d) Monotone Convergence Theorem.$

$\noindent e) The limit is the \textsl{Golden Ratio} $\varphi = \frac{1 + \sqrt{5}}{2}$.$

Thanks! Just a few queries:
(b) can we directly state that $1\leq a_n \leq 2$ for all n, even though there is a condition in (a) that x has to be between 1 and 2 (inclusive)
(c) I get the logic, but is that all I need to write for c? if not, what do I need to prove?
(d) What is the Monotone Convergence Theorem?
(e) Wow, never knew about the Golden Ratio!

InteGrand · Sep 18, 2016

1008 said:
Thanks! Just a few queries:
(b) can we directly state that $1\leq a_n \leq 2$ for all n, even though there is a condition in (a) that x has to be between 1 and 2 (inclusive)
(c) I get the logic, but is that all I need to write for c? if not, what do I need to prove?
(d) What is the Monotone Convergence Theorem?
(e) Wow, never knew about the Golden Ratio!

$\noindent b) I said what I wrote as a hint (to give the basic idea). It should be proved by induction (as the Q. says) and written up properly when you do it. The base case is trivial to check that it's true, and for the inductive step, $a_{k+1} = \sqrt{1 + a_k}\in \left[1,2\right]$, from part a) and the inductive hypothesis that $a_k \in \left[1,2\right]$. Hence by induction $a_n \in [1,2]$ for all $n$.$

$\noindent c) That was the hint for the inductive step. You'd need to write it up properly for the full proof of course (like base case etc.).$

$\noindent d) This is the crucial theorem for this sort of Q. The version we need for this Q. states that any bounded and monotonic sequence of real numbers has a limit in $\mathbb{R}$. So thanks to this theorem, we can prove a sequence has a limit if we can show it's bounded and is monotone.$

1008 · Sep 18, 2016

InteGrand said:
$\noindent b) I said what I wrote as a hint (to give the basic idea). It should be proved by induction (as the Q. says) and written up properly when you do it. The base case is trivial to check that it's true, and for the inductive step, $a_{k+1} = \sqrt{1 + a_k}\in \left[1,2\right]$, from part a) and the inductive hypothesis that $a_k \in \left[1,2\right]$. Hence by induction $a_n \in [1,2]$ for all $n$.$

$\noindent c) That was the hint for the inductive step. You'd need to write it up properly for the full proof of course (like base case etc.).$

$\noindent d) This is the crucial theorem for this sort of Q. The version we need for this Q. states that any bounded and monotonic sequence of real numbers has a limit in $\mathbb{R}$. So thanks to this theorem, we can prove a sequence has a limit if we can show it's bounded and is monotone.$

Thanks so much! Btw, how'd you know the limit in (e) was the Golden Ratio? Is it just a property of the ratio, or is there a rule/theorem attached to this?

InteGrand · Sep 18, 2016

1008 said:
Thanks so much! Btw, how'd you know the limit in (e) was the Golden Ratio? Is it just a property of the ratio, or is there a rule/theorem attached to this?

It's a famous property of the Golden Ratio (which is a famous number with many interesting properties and also appears in nature a lot apparently. See the Wikipedia page: https://en.wikipedia.org/wiki/Golden_ratio ).

I just commented it was the Golden Ratio because I thought some readers may be interested to hear that, you don't actually need to know that to do that Q. Here's how to do it (and how you should do it):

$\noindent Since $\lim _{n \to \infty} a_{n}$ exists, we can let it be $L$. Now in the equation $a_{n+1} = \sqrt{1 + a_n}$, tend $n$ to $\infty$ on both sides, so we get $L = \sqrt{1 + L}$ (the limit can go inside the square root since square root function is continuous). So $L^2 - L -1 = 0 \Rightarrow L = \frac{1+\sqrt{5}}{2}$ (taking the positive root since the sequence is positive).$

1008 · Sep 18, 2016

InteGrand said:
It's a famous property of the Golden Ratio (which is a famous number with many interesting properties and also appears in nature a lot apparently. See the Wikipedia page: https://en.wikipedia.org/wiki/Golden_ratio ).

I just commented it was the Golden Ratio because I thought some readers may be interested to hear that, you don't actually need to know that to do that Q. Here's how to do it (and how you should do it):

$\noindent Since $\lim _{n \to \infty} a_{n}$ exists, we can let it be $L$. Now in the equation $a_{n+1} = \sqrt{1 + a_n}$, tend $n$ to $\infty$ on both sides, so we get $L = \sqrt{1 + L}$ (the limit can go inside the square root since square root function is continuous). So $L^2 - L -1 = 0 \Rightarrow L = \frac{1+\sqrt{5}}{2}$ (taking the positive root since the sequence is positive).$

Thanks again, but yeah reading about the Golden Ratio, it is quite unique....
Anyway, finally get the general approach for such questions, thanks to the process just posted (I guess I was kinda overwhelmed by the trivial roots inside the roots thing that they gave at the end. I know it's an equivalent way of writing it, but it's trivial to the question)

InteGrand · Sep 18, 2016

1008 said:
Thanks again, but yeah reading about the Golden Ratio, it is quite unique....
Anyway, finally get the general approach for such questions, thanks to the process just posted (I guess I was kinda overwhelmed by the trivial roots inside the roots thing that they gave at the end. I know it's an equivalent way of writing it, but it's trivial to the question)

Maybe the reason they wrote the square roots thing is just so that people doing the Q. would realise they're finding the limit of that. Some students might not have noticed otherwise and would think they're just finding the limit of a random sequence and forget about it quickly.

I think in highschool they sometimes do Q's like this (like find value of sqrt(1+sqrt(1+…)), where the method used is call that expression x, then get x = sqrt(1+x), and solve for x). But usually in high school for these Q's, they don't prove the limit exists in the first place, which means they don't justify why we can call that expression x in the first place (so they implicitly assume without proof that it's safe to do so).

1008 · Sep 19, 2016

InteGrand said:
Maybe the reason they wrote the square roots thing is just so that people doing the Q. would realise they're finding the limit of that. Some students might not have noticed otherwise and would think they're just finding the limit of a random sequence and forget about it quickly.

I think in highschool they sometimes do Q's like this (like find value of sqrt(1+sqrt(1+…)), where the method used is call that expression x, then get x = sqrt(1+x), and solve for x. But usually in high school for these Q's, they don't prove the limit exists in the first place, which means they don't justify why we can call that expression x in the first place (so they implicitly assume without proof that it's safe to do so).

Yeah I've noticed that too, but what surprised me the most was that some of these assumptions seep into the actual HSC exam too. This is why in HSC, students that may be aware of such facts may find it logically challenging to solve problems by riding on such assumptions, where they fail to draw the line between what is safe to assume and what isn't. I guess tho HSC (3/4unit) is more about just rigorously improving your maths skills by maximising your exposure to questions so that you're able to tackle problems such as this conceptually by basing your responses on that extensive practice you've had in the past.

PS: You may be right about the reason they wrote the square roots thing.

MATH1251 Questions HELP (5 Viewers)

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