MATH1251 Questions HELP (2 Viewers)

leehuan · Oct 8, 2016

1008 said:
Thanks guys.

How'd you do this one?

I'm guessing it's really simple comparison like the one I posted above, just can think of the proper one...

$\text{A very unexpected way that my tutor employed here was to consider }\left(a_n-\frac{1}{n}\right)^2$

1008 said:
I don't think it says it's safe to assume that for this question.... Here's the proper screenshot:

Oh whoops. Ok then just stick to absolute convergence.

Paradoxica · Oct 8, 2016

leehuan said:
$\text{A very unexpected way that my tutor employed here was to consider }\left(a_n-\frac{1}{n}\right)^2$

Oh whoops. Ok then just stick to absolute convergence.

yeah that works

InteGrand · Oct 8, 2016

1008 said:
Thanks guys.

How'd you do this one?

I'm guessing it's really simple comparison like the one I posted above, just can think of the proper one...

Sorry made an error before.

$\noindent Wlog assume the $a_n$'s are non-negative (i.e. because absolute convergence implies convergence). Then by Cauchy-Schwarz, $\frac{a_1}{1} + \frac{a_2}{2} + \cdots + \frac{a_n}{n} \leq \sqrt{a_1 ^2 + a_2 ^2 + \cdots + a_n ^2}\sqrt{\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2}}$. By the Basel problem series and assumption of convergence of $\sum a_n ^2$, the RHS is bounded above by a finite number, and the LHS is thus too. Also the LHS is an increasing sequence. So the series in question has its partial sums bounded above and increasing, so it converges.$

Paradoxica · Oct 8, 2016

1008 said:
Thanks guys.

How'd you do this one?

I'm guessing it's really simple comparison like the one I posted above, just can think of the proper one...

To expand on Leehuan's comment, here's the proof:

We know that ∑(a_n)² converges, and it's well known that ∑ 1/n² converges.

By the AM-GM inequality, it follows that ∑(a_n)² + ∑ 1/n² ≥ 2∑(a_n)/n

Thus by the comparison test, it converges.

leehuan · Oct 8, 2016

I'm just trying to understand my lecture slides here.

Is this like total differential approximation?

1008 · Oct 9, 2016

leehuan said:
I'm just trying to understand my lecture slides here.

Is this like total differential approximation?

Yes, I'm pretty sure this incorporates the total differential approximation in computing the tangent plane equation to the graph of f. Notice the upside down triangle in the equation, that's the total differential approximation we're computing. It's the same as the tangent line you're used to using in 1 dimension to approximate values of a curve near the region where the line touches tangentially with the curve (i.e. the gradient of the curve = gradient of the line at that point). This idea has now been generalised to higher dimensions with planes instead of lines and surfaces instead of curves.

1008 · Oct 9, 2016

How'd you do part (d) and (e) of this question?

InteGrand · Oct 9, 2016

1008 said:
How'd you do part (d) and (e) of this question?

$\noindent d) Suppose $|z| > 3$. It is equivalent to show that $|3-z^2 + 5z^3 | < |z^5|$. Using $|z^n| = |z|^n$ and the triangle inequality, we have that$

$\begin{align*} |3-z^2 + 5z^3| &\le 3 + |z|^2 + 5|z|^3 \\ &< |z|^5, \end{align*}$

$\noindent because $3+x^2 + 5x^3 < x^5$ for all real $x$ with $x > 3$. To see why this the case, note here, for $x > 3$ (so $9 < x^2$), we have $3 + x^2 + 5x^3 < x^3 + x^3 + 5x^3 = 7x^3 < 9x^3 < x^2 \cdot x^3 = x^5$.$

$\noindent e) Suppose by way of contradiction we had a root $\alpha$ with $|\alpha|>3$. Since it's a root, we have $-\left(3\alpha -\alpha ^3 + 5\alpha^4\right) = \alpha^6$. Taking the modulus of both sides, we reach a contradiction with part e). Hence all the roots have modulus less than or equal to $3$.$

leehuan · Oct 9, 2016

InteGrand · Oct 9, 2016

leehuan said:
$\text{Show that the curve }\textbf{r}(t)=t^2\textbf{i}+t\textbf{j}+(5t-4)\textbf{k}\\ \text{is normal to the surface }2x^2+y^2+5z^2=8\\ \text{at the point }(1,1,1)$

(It is easily checked that (1,1,1) lies on the surface and on the curve.) At (1,1,1), this is t = 1 on the curve. So the curve here has tangent vector r’(1) = (2, 1, 5). A normal to the surface at (1,1,1) can be found using grad(2x^2 + y^2 + 5z^2) = (4x, 2y, 10z) to be (4,2,10), which is proportional to r’(1) as we can see. Hence r(t) curve is normal to the surface at (1,1,1).

InteGrand · Oct 9, 2016

(To get the general tangent vector r’(t), differentiate r(t) component-wise wrt t. Then sub. in t = 1 for the tangent vector at t = 1.)

leehuan · Oct 9, 2016

Ah I see. Didn't know how to deal with that "vector" function. Makes sense.
_____________________________________

$\text{Using the constant and linear terms of the Taylor polynomial in part (a), find a first order approximation to }\\ f(P+(0.01,-0.01))$

Answer to a):

$f(x,y)=\ln (4x-3y), \, P=(1,1)\\ F_3(x,y)=0+4(x-1)+3(y-1)+\frac{1}{2!}\left[-16(x-1)^2+24(x-1)(y-1)-9(y-1)^2\right]$

Also a quick question: Since a field technically satisfies the relevant axioms, it's also a vector space right?

InteGrand · Oct 9, 2016

leehuan said:
Ah I see. Didn't know how to deal with that "vector" function. Makes sense.
_____________________________________

$\text{Using the constant and linear terms of the Taylor polynomial in part (a), find a first order approximation to }\\ f(P+(0.01,-0.01))$

Answer to a):

$f(x,y)=\ln (4x-3y), \, P=(1,1)\\ F_3(x,y)=0+4(x-1)+3(y-1)+\frac{1}{2!}\left[-16(x-1)^2+24(x-1)(y-1)-9(y-1)^2\right]$

Also a quick question: Since a field technically satisfies the relevant axioms, it's also a vector space right?

Yes, any field F forms a vector space over itself (meaning that if F is a field, then taking the elements of F as the 'vectors' with the field of scalars also being F yields a vector space, where the vector addition rule is the addition rule of F and the scalar multiplication rule is the multiplication rule for elements from F).

InteGrand · Oct 9, 2016

leehuan · Oct 9, 2016

InteGrand said:
$\noindent For the Taylor approximation one, use only the up-to degree 1 terms of the Taylor polynomial you found. So it just becomes a tangent plane approximation mentioned on the previous page: $f(a+h,b+k) \approx f(a,b) + h f_{x} (a,b) + k f_{y} (a,b)$.$

Wait, is the Taylor polynomial of degree 1 essentially the equation of the tangent plane?

InteGrand · Oct 9, 2016

leehuan said:
Wait, is the Taylor polynomial of degree 1 essentially the equation of the tangent plane?

Basically you already found the Taylor polynomial about the point P in the earlier part (up to degree 2, but just take it up to degree 1 as the Q. asks for the approximation). Just sub. x = 1.01 and y = 0.99 into that polynomial to get the answer.

$\noindent And yes, the Taylor polynomial of degree 1 gives the equation of the tangent plane at the point P. I.e. in general, if $f(x,y)$ is differentiable at $(a,b)$, then its tangent plane at $(a,b)$ has equation $z = f(a,b) + f_{x} \left(a,b\right) (x-a) + f_{y} \left(a,b\right) (y-b)$, and the RHS here is the degree 1 Taylor polynomial of $f$ about $(a,b)$. It's analogous to how for a one variable function, the equation of the tangent \emph{line} at a point $a$ is $y = f(a) + f^{\prime} (a) (x-a)$, and the RHS is the degree 1 Taylor polynomial of $f$ about $a$.$

leehuan · Oct 9, 2016

$\text{A function }S:\mathbb{R}\to \mathbb{Z}\text{ is defined by }S(x)=y\text{, where }y\text{ is the integer obtained on rounding }x\text{ to the nearest integer. Is }S\text{ a linear map?}$

So I know it is not. But I was going to provide a counterexample to prove the map's bound to fail the addition condition somehow. But the answers say that Z is not a vector space - which axiom is it breaking?

InteGrand · Oct 9, 2016

leehuan said:
$\text{A function }S:\mathbb{R}\to \mathbb{Z}\text{ is defined by }S(x)=y\text{, where }y\text{ is the integer obtained on rounding }x\text{ to the nearest integer. Is }S\text{ a linear map?}$

So I know it is not. But I was going to provide a counterexample to prove the map's bound to fail the addition condition somehow. But the answers say that Z is not a vector space - which axiom is it breaking?

For one thing, the codomain isn't a vector space. So it's not a linear map, since linear maps are generally defined from one vector space to another.

Anyway, one way to show it fails to preserve addition (it is the floor function):

floor(1.5 + 1.5) = floor(3) = 3 =/= floor(1.5) + floor(1.5) = 1 + 1 = 2.

leehuan · Oct 9, 2016

InteGrand said:
For one thing, the codomain isn't a vector space. So it's not a linear map, since linear maps are defined from one vector space to another.

Except I thought Z was. But I just realised something

$\text{Did I confuse }(\mathbb{Z}, +, \cdot, \mathbb{R})\text{ with }(\mathbb{Z}, +, \cdot, \mathbb{Z})?$

InteGrand said:
Anyway, one way to show it fails to preserve addition (it is the floor function):

floor(1.5 + 1.5) = floor(3) = 3 =/= floor(1.5) + floor(1.5) = 1 + 1 = 2.

Yeah sweet that's what I had in mind

InteGrand · Oct 9, 2016

leehuan said:
Except I thought Z was. But I just realised something

$\text{Did I confuse }(\mathbb{Z}, +, \cdot, \mathbb{R})\text{ with }(\mathbb{Z}, +, \cdot, \mathbb{Z})?$

Z isn't a field (can't in general find multiplicative inverses in Z, e.g. 2 doesn't have a multiplicative inverse in Z), so we can't call it a vector space over itself like we can with say R (or general fields F).

Z equipped with a field of scalars R still wouldn't be a vector space, with ordinary multiplication, since αn is not always an integer for α real and n integer, e.g. take α = 1/2 and n = 1 (i.e. we wouldn't have closure under scalar multiplication).

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