Calculus & Analysis Marathon & Questions (3 Viewers)

leehuan · Apr 24, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
An equation can still have infinite solutions.

For this question, because infinity is not a measurable quantity (and I'm not sure if you can assume that it is an integer) we can't say for sure what the behaviour of f is if f(x)=y yields an infinite number of solutions.

At least, my simple brain can't visualise f anymore

porcupinetree · Apr 24, 2016

Re: First Year Uni Calculus Marathon

leehuan said:
$$We first request that $f$ is continuous at $x=0$

$\\ \lim_{x\rightarrow 0^+}f(x) = \lim_{x\rightarrow 0^+}{x}^{3} =0\\\lim_{x\rightarrow 0^-}f(x) = \lim_{x\rightarrow 0^=}{x}^{3} =0 \\ \therefore \lim_{x\rightarrow 0}f(x) =0$

$\\ $So since $\lim_{x\rightarrow 0}f(x) = f(0) = 0\\ f(x)$ is indeed continuous at 0$$

$The derivative of a piecewise function cannot be computed by immediately taking derivatives at the point where $f$ changes it's rule.$

$\\ \lim _{ h\rightarrow { 0 }^{ + } }{ \frac { f\left( 0+h \right) -f\left( 0 \right) }{ h } } =\lim _{ h\rightarrow { 0 }^{ + } }{ \frac { { h }^{ 3 }-{ 0 }^{ 3 } }{ h } } =\lim _{ h\rightarrow { 0 }^{ + } }{ { h }^{ 2 } } =0\\ \lim _{ h\rightarrow { 0 }^{ - } }{ \frac { f\left( 0+h \right) -f\left( 0 \right) }{ h } } =\lim _{ h\rightarrow { 0 }^{ - } }{ \frac { { h }^{ 3 }-{ 0 }^{ 3 } }{ h } } =\lim _{ h\rightarrow { 0 }^{ - } }{ { h }^{ 2 } } =0\\ \therefore \lim _{ h\rightarrow { 0 } }{ \frac { f\left( 0+h \right) -f\left( 0 \right) }{ h } } =f^{ { \prime } }\left( 0 \right) =0$

$Hence, as the slope of the tangent is similar from both sides, $f$ must also be differentiable at 0 (or in other words, $f^\prime (0)$ is defined)$
=======================
$Next question$

$$Determine the behaviour of $f(x)=\frac{\ln{\left(x^3+1\right)}}{\ln{\left(x^2+1\right)}}$ as $x\rightarrow \infty$

$Add on question: Show that the relevant definition is satisfied.$

Also, regarding your proof, I think you need to the read the question again. (Particularly the definition of f(x))

leehuan · Apr 24, 2016

Re: First Year Uni Calculus Marathon

Not even sure what I was thinking at the time now

Paradoxica · Apr 24, 2016

Re: First Year Uni Calculus Marathon

porcupinetree said:
Part 1 (not a very rigorous proof, though):

$$As $x\rightarrow \infty $, $x^3 + 1 \rightarrow x^3$
$$As $x\rightarrow \infty $, $x^2 + 1 \rightarrow x^2$
$$Thus, $ \lim_{x \rightarrow \infty} \frac{\ln (x^3 + 1)}{\ln (x^2 + 1)} = \frac{\ln(x^3)}{\ln(x^2)} = \frac{3\ln(x)}{2\ln(x)} = \frac{3}{2}$

Here's my attempt:

Since x is positive, we can invoke the following inequalities:

$$\noindent$ x^2 < x^2+1 < x^2 + 2x + 1 \\ x^3 < x^3 + 1 < x^3 + 3x^2 + 3x +1$

$$\noindent After some muddling around, we get the following:$ \\ \frac{3\log{x}}{2\log{(x+1)}} < L < \frac{3\log{(x+1)}}{2\log{x}}$

It is easy to show that the left hand side is bounded from above by 3/2, and the right hand side is bounded from below by the same value.

It is then sufficient to show the left bound is monotonically increasing, and the right bound is monotonically decreasing, for sufficiently large x.

leehuan · Apr 24, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
Here's my attempt:

Since x is positive, we can invoke the following inequalities:

$$\noindent$ x^2 < x^2+1 < x^2 + 2x + 1 \\ x^3 < x^3 + 1 < x^3 + 3x^2 + 3x +1$

$$\noindent After some muddling around, we get the following:$ \\ \frac{3\log{x}}{2\log{(x+1)}} < L < \frac{3\log{(x+1)}}{2\log{x}}$

It is easy to show that the left hand side is bounded from above by 3/2, and the right hand side is bounded from below by the same value.

It is then sufficient to show the left bound is monotonically increasing, and the right bound is monotonically decreasing, for sufficiently large x.

Believe it or not my workbook was happy if you just used L'Hopitals

Paradoxica · Apr 24, 2016

Re: First Year Uni Calculus Marathon

leehuan said:
Believe it or not my workbook was happy if you just used L'Hopitals

Not rigorous enough

jks. But I cringe at directly using that on the expression, I would rather use it on the bounds, it's much more nice-looking.

leehuan · Apr 24, 2016

Re: First Year Uni Calculus Marathon

@IG

Still suspecting I haven't fully finished seanieg's question though. Anything else to pick up?

InteGrand · Apr 24, 2016

Re: First Year Uni Calculus Marathon

^ I think you should try invoking the IVT to prove some things I think you're trying to prove.

seanieg89 · Apr 24, 2016

Re: First Year Uni Calculus Marathon

(Sorry about delayed response, am wrestling with a problem of my own

)

Yep @leehuan, IVT, and EVT are key to rigorous argument here. You are on a nice train of thought and I think you have a good intuitive grasp of why k=2 is impossible, but making this rigorous takes some care.

The "infinitesimal" part of your proof in particular needs to be made more precise, and things like Rolle's theorem are inapplicable here because the function need not be differentiable.

From what you are saying though, it seems to me you have the correct "picture" in your head for your k=2 disproof.

(I am also unconvinced that this "picture" serves to disprove the k>2 possibility so think about explaining that a little clearer).

seanieg89 · Apr 24, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
But what about k = ∞ ?

Yes it is possible. Eg: f(x)=x*sin(x).

porcupinetree said:
Part 1 (not a very rigorous proof, though):

$$As $x\rightarrow \infty $, $x^3 + 1 \rightarrow x^3$
$$As $x\rightarrow \infty $, $x^2 + 1 \rightarrow x^2$
$$Thus, $ \lim_{x \rightarrow \infty} \frac{\ln (x^3 + 1)}{\ln (x^2 + 1)} = \frac{\ln(x^3)}{\ln(x^2)} = \frac{3\ln(x)}{2\ln(x)} = \frac{3}{2}$

This is easy to make rigorous. For large x, x^3+1 < 2x^3 and x^2+1 < 2x^2. So

log(x^3+1)/log(x^2+1) < log(2x^3)/log(x^2) = (log(2)+3log(x))/2log(x).

and

log(x^3+1)/log(x^2+1) > log(x^3)/log(2x^2)=3log(x)/(log(2)+2log(x)).

Squeeze to finish.

leehuan said:
Hmm...

$$Cases where differentiability arise out of a discontinuity can be safely disregarded as the question already specified $f$ is continuous on $\mathbb R$

$This means that the lack of differentiability arises in the event of a cusp or a corner. However, that same cusp/corner will essentially also be a local extrema, thus the proof proceeds from there in a similar manner.$

At least, I don't think a cusp/corner can't be a local extrema here?

Corners do not have to be local extrema. Eg f(x)=max(x,2x) at 0.

Also, lack of differentiability can occur in ways worse than corners. (f(x)-f(a))/(x-a) needs not tend to a limit at all as x tends to a, or it might only tend to a limit from one side.

It is more trouble than it is worth to appeal to theorems about differentiable functions and then treat the exceptional points separately. (In fact a continuous function might be differentiable nowhere!)

leehuan · Apr 24, 2016

Re: First Year Uni Calculus Marathon

Hmmm

With the IVT I can pretty much find how to apply it. Just get rid of that infinitesimal stuff and properly redefine it (which I might do later). Which gives me some ideas on the EVT but I haven't placed too much thought into it to know what's right and what's flawed.

May continue this when I have more free time. Gah I hate assignment based subjects

Paradoxica · Apr 24, 2016

Re: First Year Uni Calculus Marathon

seanieg89 said:
Yes it is possible. Eg: f(x)=x*sin(x).

This is easy to make rigorous. For large x, x^3+1 < 2x^3 and x^2+1 < 2x^2. So

log(x^3+1)/log(x^2+1) < log(2x^3)/log(x^2) = (log(2)+3log(x))/2log(x).

and

log(x^3+1)/log(x^2+1) > log(x^3)/log(2x^2)=3log(x)/(log(2)+2log(x)).

Squeeze to finish.

Corners do not have to be local extrema. Eg f(x)=max(x,2x) at 0.

Also, lack of differentiability can occur in ways worse than corners. (f(x)-f(a))/(x-a) needs not tend to a limit at all as x tends to a, or it might only tend to a limit from one side.

It is more trouble than it is worth to appeal to theorems about differentiable functions and then treat the exceptional points separately. (In fact a continuous function might be differentiable nowhere!)

*cough*weierstrass*cough*

InteGrand · Apr 24, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
*cough*weierstrass*cough*

Imagine Carrotsticks BOS trial last Q. of 4U paper 2016: 'Sketch the graphs of the Weierstrass and Dirichlet functions'

.

Paradoxica · Apr 24, 2016

Re: First Year Uni Calculus Marathon

InteGrand said:
Imagine Carrotsticks BOS trial last Q. of 4U paper 2016: 'Sketch the graphs of the Weierstrass and Dirichlet functions' .

last part of the question

"also note significant points on the graph"

Paradoxica · Apr 24, 2016

Re: First Year Uni Calculus Marathon

InteGrand said:
Imagine Carrotsticks BOS trial last Q. of 4U paper 2016: 'Sketch the graphs of the Weierstrass and Dirichlet functions' .

the final part will be:

hence, sketch the indefinite integral of the weierstrass and dirichlet functions, with any choice of C you like.

leehuan · Apr 25, 2016

Re: First Year Uni Calculus Marathon

porcupinetree said:
$Next question:$
$Show that $f$, where $f$ is defined by$
$f(x) = \left\{\begin{matrix} x^3 & $if $x$ is rational$ \\ 0 & $if $x$ is irrational$ \end{matrix}\right.$
$is differentiable at $x=0$.$

Regarding the continuity part:

Informally, when x≥0, 0≤f(x)≤x³ and when x≤0, x³≤f(x)≤0

But in both cases if we take limits to nought, by the squeeze theorem we have lim x->0 f(x) = 0 which incidentally enough equals f(0)
______________________

The answer given to the derivative was wrong because I didn't regard the significance of what h was.

lim x->0 [ f(0+h)-f(0) ] / h
= lim x->0 f(h)/h

but f(h) = h³ when h is rational
and f(h) = 0 when h is irrational

Note then that f(h)/h = h² for h in Q
or = 0 for h not in Q

Haven't done questions like this one before yet but just staring at it I'm inclined to say the squeeze theorem can be used again

InteGrand · Apr 25, 2016

Re: First Year Uni Calculus Marathon

Yeah it's basically squeeze law. We can use that to show that f'(0) = 0.

seanieg89 · Apr 25, 2016

Re: First Year Uni Calculus Marathon

leehuan said:
Haven't done questions like this one before yet but just staring at it I'm inclined to say the squeeze theorem can be used again

$\left|\frac{f(x)-f(0)}{x-0}\right|=\frac{|f(x)|}{|x|}\leq \frac{|x|^3}{|x|}=|x|^2\rightarrow 0\\ \\ \Rightarrow \frac{f(x)-f(0)}{x-0}\rightarrow 0.$

(The "=>" comes from the squeeze law. Generally if we suspect something might tend to zero, it is often convenient to take absolute values as then we just need to bound it above by something that tends to zero.)

porcupinetree · Apr 25, 2016

Re: First Year Uni Calculus Marathon

$\lim_{x \rightarrow 0} \left(\frac{1+\tan{x}}{1+\sin{x}}\right)^{\frac{1}{\sin{x}}}$

leehuan · Apr 25, 2016

Re: First Year Uni Calculus Marathon

porcupinetree said:
$\lim_{x \rightarrow 0} \left(\frac{1+\tan{x}}{1+\sin{x}}\right)^{\frac{1}{\sin{x}}}$

$\lim _{ x\rightarrow 0 } \left( \frac { 1+\tan { x } }{ 1+\sin { x } } \right) ^{ \frac { 1 }{ \sin { x } } }\\ =\lim _{ x\rightarrow 0 }{ { e }^{ \ln { { \left( \frac { 1+\tan { x } }{ 1+\sin { x } } \right) } } ^{ \csc { x } } } } \\ =\lim _{ x\rightarrow 0 }{ { e }^{ \frac { \ln { \left( 1+\tan { x } \right) } -\ln { \left( 1+\sin { x } \right) } }{ \sin { x } } } } \\ ={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { \ln { \left( 1+\tan { x } \right) } -\ln { \left( 1+\sin { x } \right) } }{ \sin { x } } } }\\ \stackrel{L'H}{=} { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \frac { \sec ^{ 2 }{ x } }{ 1+\tan { x } } -\frac { \cos { x } }{ 1+\sin { x } } }{ \cos { x } } } }\\ ={ e }^{ \frac { 0 }{ 1 } }\\ =1$

$Use of L'H was justified as the final limit exists.$

Tightly squeezed LaTeX 101

Calculus & Analysis Marathon & Questions (3 Viewers)

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