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[WSG23] Daily Study Group: Introduction to Calculus

A Wolfram U Daily Study Group on "Introduction to Calculus" begins on Monday, April 17, 2023.

Join a cohort of fellow mathematics enthusiasts to learn about the fundamentals of calculus along with its applications from video lessons created for the popular Introduction to Calculus course. Participate in live Q&A and review your understanding through interactive in-session polls. Complete quizzes at the end of the study group to get your certificate of program completion.

The study group will be led by expert Wolfram U instructor Luke Titus and should be great fun!

April 17- May 5, 2023, 11am-12pm CT (4-5pm GMT)

REGISTER HERE

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Please feel free to use this thread to collaborate and share ideas, materials and links to other resources with fellow learners.
POSTED BY: Devendra Kapadia
23 Replies
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Phil Earnhardt
Phil Earnhardt
Posted 2 years ago
POSTED BY: Phil Earnhardt
Phil Earnhardt
Phil Earnhardt
Posted 2 years ago
POSTED BY: Phil Earnhardt

5-4-23
Dear Professor Kapadia,
Great course. I'm a newbie and learning. About Riemann integrability, I read that Lebesgue integrability is also a possibility if the Lebesgue measure is 0.
This then allows Riemann integrability and this justifies a closed interval, but with a Weierstrass covering. The points at which there is no differentiability are allowed f they have Lebesgue measure 0. But can the genus be above 0 ever.
Apologies if my terminology is a bit off.
My basic question is whether Lesbesgue integrability is of use beyond Riemann integrals?
Thanks,
Paul Shapshak, PhD
USF
POSTED BY: Paul Shapshak, PhD
Phil Earnhardt
Phil Earnhardt
Posted 2 years ago
POSTED BY: Phil Earnhardt
POSTED BY: Luke Titus
Phil Earnhardt
Phil Earnhardt
Posted 2 years ago
POSTED BY: Phil Earnhardt

Thank you for sharing that video, Phil. That is very interesting. I've seen a lot of ways to calculate Pi in situations where it would seem a circle or Pi shouldn't appear, but this is certainly one of the most intriguing ones.

POSTED BY: Luke Titus
POSTED BY: Zbigniew Kabala

The limit doesn't exist at x=1, so you can't calculate the derivative as that involves a limit at that point. 

In[1]:= f[x_] := Piecewise[{{x^2, x <= 1}, {x^2 + 2, x > 1}}]
Limit[f[x], x -> 1]

Out[2]= Indeterminate
POSTED BY: Luke Titus
POSTED BY: Zbigniew Kabala
POSTED BY: Luke Titus

POSTED BY: Zbigniew Kabala
POSTED BY: Luke Titus

POSTED BY: Zbigniew Kabala
POSTED BY: Luke Titus

In the following expression given in the exercises, the function IS defined at x=-3 because the piecewise condition says x greater than or equal to -3. Yet the Limit function says it is indeterminate at x=-3. If it had said x>-3 I get it. The function was never defined at x=-3.

What gives?

f[x_]:=Piecewise[{{x-5,x<-3},{6/(x+4),x>=-3}}]
POSTED BY: Carl Hahn
Phil Earnhardt
Phil Earnhardt
Posted 2 years ago

The Limit[f[x],x->=-3] is indeterminate, because different values are derived when the limit is approached from above or from below.

However, you can do 

Limit[f[x],x->-3,Direction->"FromAbove"]

and

Limit[f[x],x->-3,Direction->"FromBelow"]

to specify the limits from each side.
POSTED BY: Phil Earnhardt
POSTED BY: Carl Hahn
Phil Earnhardt
Phil Earnhardt
Posted 2 years ago

The limit to f[x], approached from above, is 6

The limit to f[x], approached from below, is -8

You can see those 2 limits computed by using that option with the Limit[] function. If you do not specify fromabove/frombelow , my understanding is that the limit is indeterminate if it doesn't approach the same value from the 2 sides of the limit.

This example is showing the difference between evaluating at the value at a location, and determining the limit as it approaches that value. That is a difference -- the crucial difference -- between those two things. I believe that's the case, but I don't see in the Wolfram Language documentation where the behavior of Limit[] is formally stated.

POSTED BY: Phil Earnhardt
Phil Earnhardt
Phil Earnhardt
Posted 2 years ago
POSTED BY: Phil Earnhardt

In general, the limit of a function at 'a' has nothing to do with its value at 'a', unless the function is continuous at 'a', which happens when the value at 'a' coincides with its limit at 'a'. In the limit, you only care about values AROUND 'a', different than 'a'. The limit may NOT exist at 'a', even if f(a) is defined and the limit at 'a' may exist even if f(a) is not defined.
POSTED BY: Juan Ortiz Navarro
Phil Earnhardt
Phil Earnhardt
Posted 2 years ago
POSTED BY: Phil Earnhardt

Hi Phil! It's great to have you in the study group. I'm glad you enjoyed some of the earlier ones that Arben taught. What you are working on with impedance sounds very interesting. I hope you enjoy the study group.
POSTED BY: Luke Titus
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