Calculus & Analysis Marathon & Questions (10 Viewers)

integral95 · Jun 7, 2016

Re: First Year Uni Calculus Marathon

leehuan said:
Yeah basically, but part a) could've been smacked straight away with the first FTC and the chain rule

Hiccup on the final answer for d) though

wait what?
I don't know if it's legit lol

leehuan · Jun 7, 2016

Re: First Year Uni Calculus Marathon

integral95 said:
wait what?
I don't know if it's legit lol

f'(1) = 3√17 not 3√5

InteGrand · Jun 12, 2016

Re: First Year Uni Calculus Marathon

$\noindent Let $P$ and $Q$ be non-constant polynomials with positive leading coefficients. Show that $\lim _{y\to \infty}\frac{P\left(y\right)}{Q\left(\ln y\right)}=\infty$.$

leehuan · Jun 12, 2016

Re: First Year Uni Calculus Marathon

I'm taking a massively blank stab at this one for my procrastination from accounting, so the quality of this answer will be again, poor lol.

$$If $\deg{P}\ge \deg{Q}$, then $\\L=\lim_{y \to \infty}{\frac{a_{n+a}y^{n+a}+a_{n+a-1}y^{n+a-1}+\dots+a_ny^n+a_{n-1}y^{n-1}+\dots+a_0}{b_n{(\ln{y}) }^n + b_{n-1}(\ln{y}) ^{n-1} + \dots + b_0 } }$

$$Then, just the fact that as $x$ tends to $\infty, \\ \ln{x}=o(x)$ and consequently $(\ln{x})^n=o\left(x^n\right)$ can be used to show that the numerator increases faster than the denominator, i.e. the limit diverges.$\\ $Note: Obviously $ y^{n+a}>y^n $ for large $y$ if $n \ge 1$

$$If $\deg{P} < \deg{Q}$ however, note that $y^m \to \infty, (\ln{y})^n \to \infty$ as $ y \to \infty$ so we can use L'Hopital's rule (for any $m>0, n>0)$

$\begin{align*} \lim_{y \to \infty}{\frac{P(y)}{Q(\ln{y})}} &\stackrel{L'H}{=} \lim_{y \to \infty}{\frac{P^\prime (y)}{Q^\prime (\ln{y}) \cdot \frac{1}{y} } } \\ &= \lim_{y \to \infty}{\frac{y P^\prime (y)}{Q^\prime (\ln{y}) } } \\ &\stackrel{L'H}{=} \lim_{y \to \infty}{\frac{y P^{\prime \prime} (y) + P^\prime (y)}{Q^{\prime \prime} (\ln{y}) \cdot \frac{1}{y} } } \\ &= \lim_{y \to \infty}{\frac{y^2 P^{\prime \prime} (y) + y P^\prime (y)}{Q^{\prime \prime} (\ln{y}) } } \\ &\stackrel{L'H}{=} \lim_{y \to \infty}{\frac{y^3 P^{\prime \prime \prime} (y) +3y^2 P^{\prime \prime} (y) + y P^\prime (y)}{Q^{\prime \prime \prime} (\ln{y}) } } \end{align*}\\ \text{and so on}$

$Interestingly, with each application of L'Hopital's rule, the degree of the numerator stays the same, but the degree of the denominator decreases by 1. This is because the $\frac{1}{y}$ introduced into the denominator as per each application of L'H raises the degree of the numerator back up by one, each time that decreases (hence, overall effects cancel out). After a certain amount of derivatives being taken, the denominator will become a constant term, which will obviously imply the limit tends to infinity.$

dan964 · Jun 13, 2016

Re: First Year Uni Calculus Marathon

Wouldn't you have to also check/assume the individual requirements of L'Hs are satisfied, namely the limits have to all exist?

Paradoxica · Jun 13, 2016

Re: First Year Uni Calculus Marathon

InteGrand said:
$\noindent Let $P$ and $Q$ be non-constant polynomials with positive leading coefficients. Show that $\lim _{y\to \infty}\frac{P\left(y\right)}{Q\left(\ln y\right)}=\infty$.$

Change variables.

P(e^x)/Q(x)

Without Loss of Generality, we can assume P is of degree 1.

This expression obviously tends to infinity.

For those who do not see it, apply L'Hôpital's Rule (with respect to x) n times where n is the degree of Q.

If P is larger, then the contribution to the explosion is greater, and the expression still diverges.

[QED]

BlueGas · Jun 13, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
Change variables.

P(e^x)/Q(x)

Without Loss of Generality, we can assume P is of degree 1.

This expression obviously tends to infinity.

For those who do not see it, apply L'Hôpital's Rule (with respect to x) n times where n is the degree of Q.

If P is larger, then the contribution to the explosion is greater, and the expression still diverges.

[QED]

How are you able to do these uni questions when you're not even at uni?

Paradoxica · Jun 13, 2016

Re: First Year Uni Calculus Marathon

BlueGas said:
How are you able to do these uni questions when you're not even at uni?

These things are fairly simple to understand, you don't have to be taking a high level mathematics course to be able to do them. I'm sure I could find people in year 10 who can do these questions.

leehuan · Jun 13, 2016

Re: First Year Uni Calculus Marathon

dan964 said:
Wouldn't you have to also check/assume the individual requirements of L'Hs are satisfied, namely the limits have to all exist?

That's a statement I forgot to (explicitly) make.

But I obviously assumed so because the final limit is supposed to exist and tend to infinity

BlueGas · Jun 13, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
These things are fairly simple to understand, you don't have to be taking a high level mathematics course to be able to do them. I'm sure I could find people in year 10 who can do these questions.

That's abit of an exaggeration isn't it? Would you find people in Year 10 (unless if they accelerate) that understand L'Hôpital's Rule?

Paradoxica · Jun 13, 2016

Re: First Year Uni Calculus Marathon

BlueGas said:
That's abit of an exaggeration isn't it? Would you find people in Year 10 (unless if they accelerate) that understand L'Hôpital's Rule?

Yes. I would, and I can, and I will. I am quasi-mentoring (mainly because he is very self capable) a year 10 student who has self-taught all the basics of Extension 2 Integration, and I will be teaching him competitive integration. Then I will drag him to the MATHSOC Integration Bee next year.

dan964 · Jun 13, 2016

Re: First Year Uni Calculus Marathon

leehuan said:
That's a statement I forgot to (explicitly) make.

But I obviously assumed so because the final limit is supposed to exist and tend to infinity

Limits that tend to infinity, do they actually exist in the reals?

InteGrand · Jun 13, 2016

Re: First Year Uni Calculus Marathon

Going to infinity counts as "exist" for the purposes of using L'Hôpital's rule. You can't have something that oscillates forever though (like (2+cos(x))/(2-cos(x)) or something), that'd fail to satisfy the requirement for L'Hôpital's rule.

Paradoxica · Jun 13, 2016

Re: First Year Uni Calculus Marathon

InteGrand said:
Going to infinity counts as "exist" for the purposes of using L'Hôpital's rule. You can't have something that oscillates forever though (like (2+cos(x))/(2-cos(x)) or something), that'd fail to satisfy the requirement for L'Hôpital's rule.

Identically equal to making a manipulation/substitution so that you get 0/0 instead of ∞/∞

Paradoxica · Jun 26, 2016

Re: First Year Uni Calculus Marathon

$f(x) = \begin{cases} \tan^{-1}{x} \quad \text{if } x \text{ is rational}\\ \sin{x} \quad \quad \, \text{if }x\text{ is irrational} \end{cases}$

Prove that f is differentiable at the origin.

There is a real positive number θ which satisfies tan^-1θ = sinθ.

Prove that f is continuous at θ.

Hence prove f is continuous at -θ.

leehuan · Jun 26, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
Yes. I would, and I can, and I will. I am quasi-mentoring (mainly because he is very self capable) a year 10 student who has self-taught all the basics of Extension 2 Integration, and I will be teaching him competitive integration. Then I will drag him to the MATHSOC Integration Bee next year.

...just saw (realised) this

Paradoxica · Jun 26, 2016

Re: First Year Uni Calculus Marathon

leehuan said:
...just saw (realised) this

cool

InteGrand · Jul 2, 2016

Re: First Year Uni Calculus Marathon

$$\noindent Let $f\colon\mathbb{R}\to \mathbb{R}$ be a function that is continuous on an open interval $I$ containing $a\in \mathbb{R}$ and is differentiable on $I$, except possible at $a$. Suppose $\lim_{x\to a}f^{\prime}\left(x\right)$ exists. Does $f^\prime \left(a\right)$ necessarily exist? Justify your answer.$

seanieg89 · Jul 2, 2016

Re: First Year Uni Calculus Marathon

InteGrand said:
$$\noindent Let $f\colon\mathbb{R}\to \mathbb{R}$ be a function that is continuous on an open interval $I$ containing $a\in \mathbb{R}$ and is differentiable on $I$, except possible at $a$. Suppose $\lim_{x\to a}f^{\prime}\left(x\right)$ exists. Does $f^\prime \left(a\right)$ necessarily exist? Justify your answer.$

Yep. MVT implies (f(x)-f(a))/(x-a)=f'(c(x)) for some c(x) strictly between a and x. As x->a, c(x)->a, which implies that the RHS converges to a limit by the question's assumptions. So the LHS converges to a limit as x->a, which is precisely the definition of differentiability.

A related followup:
Let f:R->R be a differentiable function. Must f' be continuous?

leehuan · Jul 2, 2016

Re: First Year Uni Calculus Marathon

seanieg89 said:
Yep. MVT implies (f(x)-f(a))/(x-a)=f'(c(x)) for some c(x) strictly between a and x. As x->a, c(x)->a, which implies that the RHS converges to a limit by the question's assumptions. So the LHS converges to a limit as x->a, which is precisely the definition of differentiability.

A related followup:
Let f:R->R be a differentiable function. Must f' be continuous?

Going by memory isn't the function f(x)=x. sin(1/x) for x \neq 0 ; 0 for x=0 a case where f is differentiable everywhere but f' is not continuous at 0?

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