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Re: HSC 2016 4U Marathon Yep (that's a sufficient condition, but not necessary). This follows from similar triangles.

Re: HSC 2016 4U Marathon $\noindent You can (and should) try checking your answer with easy values of $\alpha,\beta$. Note that if $\alpha = 2$ and $\beta = 1$, the inequality in Paradoxica's post holds but the one in yours doesn't.$ (Edit: I think you edited your inequality. Or I probably...

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...

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) =...

Re: First Year Uni Calculus Marathon Is this inequality basically an application of (continuous version of) Jensen's inequality?

Yeah, you can.

I think Mystery Mark means under 30 ATAR. (If you get that, apparently your ATAR just gets reported to you as "*", and you don't get told your actual numerical ATAR.)

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...

$\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: $f(a+h,b+k) \approx f(a,b) + h f_{x} (a,b) + k f_{y} (a,b)$.$

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...

(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.)

(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...

$\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...

(Those relationships hold true for the solutions to the normal equations for fitting to a line of the form y = mx + b.)

Yes

Well done! :)

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} +...

$\noindent To use the comparison test, we'd need the summands to have non-negative terms. We can get around this by showing the desired sum converges absolutely though, because then the series we'd consider would have positive terms. I.e. note $c_n := |a_n b_n| \leq \frac{a_n ^2 + b_n^2}{2}$ for...

They are the same. Edit: said above.

$\noindent Consider the sequence of numbers $F_{1},F_{2},F_{3},\ldots$, defined recursively by $F_{n} = F_{n-1} + F_{n-2}$ for $n \geq 3$, and $F_{1} = F_{2} = 1$. (This is the \textit{Fibonacci sequence}.)$ $\noindent (i) Use induction to show that for any real number $x$ satisfying $x^2 - x...