Since 1/|1+x^2| is bounded above by 1 and |x|/|1+x^2| is bounded above by 1/2 (and similarly for y), will the proof still hold if we had chosen delta = epsilon?
Why is the last inequality "less than or equal to 1" instead of strictly less than 1? What I mean is, does there exists a number a such that (|a| / |1+a^2| )= 1?
mr math sorcerer, at 7:30 can we prove this segment by proving the bottom part of the fraction is bigger or equal to the top part? what i mean by this is by plugging in any number for 'a' we get a result smaller or equal to one
could just let delta equal n* epsilon at first then when he got to the end of proof just change n to 1/2 as that clearly works nicely with the mod(f(x)-f(y)) being less than epsilon
In the proof at the end, shouldn't you be considering the cases where |a| 1? The way you did it means that if a < -1, then |a| > 1 and you can't use the first inequality in case 1.
Matt Hutchings oh yeah! I get your point. Those cases work only for non-negative numbers. For the whole real line we have to go with cases where a is between -1 & 1 and otherwise. Great observation.
immediately frustrated because you figured out delta beforehand and was just like ok there you go. now let's do more redundant stuff to see if it's actually what I put down though I already told you it is.
Exactly. Finding the delta is the actual problem. Not showing that the chosen delta works. I fuckin despise these delta epsilon proofs, they make me go insane.
I'm confused. Why is it redundant...? You won't be given delta in an actual problem.. He just said, "this is what I came up with," and showed us how he came up with it.
The absolute value of the difference of real numbers is a metric on the real numbers, so you may change the order of the terms and obtain an equivalent expression. More simply, notice that the absolute value of a difference gives the distance on a number line between the two values. This distance shouldn't change just because we change the order of the values. Ex. 4 - 9 = -5, 9 - 4 = 5, but |-5|=|5| = 5
Well, since it is absolute value it must be positive. And of if it is positive it can only makes that whole fraction smaller number. So if you get rid of it it doesnt really matter.
Ivan Ivan It indeed becomes a smaller number, but we claim that is is greater of equal, right? So how can two numbers suddenly be greater when they become smaller?
+The Math Sorcerer um if memory serves the rule I worked out was to check what is the largest number the 1/(1+x^2)+1/(1+y^2) could be which is 2 in this case the dive epsilon by that number and call it delta.
I am sooo late I have a doubt...i hope you reply.....i understand the proofs...but is there any tips on determining the delta......i understand it depends on the question but some tips on that
because it's absolute value we know the things he is "dropping" are positive and we also know that the bigger the denominator of a number the smaller the number. So if we're working with inequalities it follows that |x|/ |1+x^2| |1+y^2| < |x|/ |1+x^2|
+James Digno a^2>=0 for all a in R, so (1+a^2)>=(1) => mod(1+a^2)>=mod(1) thus dividing by the smaller expression will give a greater expression as it is a reciprocal (laws of inequalities). A similiar argument will convince you that dropping the 1 will also give the same result.
Great video but please explain how is this function uniformly continuous if when I graph it it does not have a constant slope (isn't that required to be uniformly continuous)
I still can't see the difference between continuity and uniform continuity. I think I don't understand them correctly. Please please please please please make video on them and show the differences
I just wanna say thank you, your videos not only explain everything in detail but also saved me a huge amount of time.
it will probably take me years before this whole analysis course make sense
@@markdierker4292 about to graduate without fully knowing proofs
agree
@@lemyul did you figure out proofs yet?
Since 1/|1+x^2| is bounded above by 1 and |x|/|1+x^2| is bounded above by 1/2 (and similarly for y), will the proof still hold if we had chosen delta = epsilon?
The best 4 minutes I ever spent (2x speed works like a dream)
Love your proves! Keep it on dude, you are saving lot of asses around the world
haha thanks
There is a problem with the yellow proof at the end. Consider a=-5. -5
Why is the last inequality "less than or equal to 1" instead of strictly less than 1? What I mean is, does there exists a number a such that (|a| / |1+a^2| )= 1?
mr math sorcerer, at 7:30 can we prove this segment by proving the bottom part of the fraction is bigger or equal to the top part?
what i mean by this is by plugging in any number for 'a' we get a result smaller
or equal to one
So, if I were cluelessly doing this, I could just choose 𝜺 = 𝛿 and in the end see that the |f(x)-f(y)|
how do you know that the delta value is epsilon/2?
it finally all makes sense thank you
np happy it helped:)
This is soooooo helpful. Thank you so much!
You're so welcome!
Awesome video, I really liked the way you make the proof at the end of it about 1+1.
thanks:)
How the heck did you just arbitrarily drop abs(1+ y^2) from the bottom of the fraction?
when abs(1+ y^2) drops, all of the expression gets bigger so no problem to drop it eheh
Where and how did you get delta equals to epsilon over 2?
could just let delta equal n* epsilon at first then when he got to the end of proof just change n to 1/2 as that clearly works nicely with the mod(f(x)-f(y)) being less than epsilon
In the proof at the end, shouldn't you be considering the cases where |a| 1? The way you did it means that if a < -1, then |a| > 1 and you can't use the first inequality in case 1.
Matt Hutchings oh yeah! I get your point. Those cases work only for non-negative numbers. For the whole real line we have to go with cases where a is between -1 & 1 and otherwise. Great observation.
any examples of determining if a function is continuous and or discontinuous?
Thanks professor for your videos. They help us a lot. With due respect, your proof for the lemma used is wrong(for the case when a
Why do you choose delta=epsilon/2?
Only after completing the problem , we can take this. This is one example for how mathematicians do Mathematics.
Nice, you could've also shown x/1+x^2 as a sin function and hence
immediately frustrated because you figured out delta beforehand and was just like ok there you go. now let's do more redundant stuff to see if it's actually what I put down though I already told you it is.
Exactly. Finding the delta is the actual problem. Not showing that the chosen delta works. I fuckin despise these delta epsilon proofs, they make me go insane.
It's always between 0 and epsilon so in 99% of the cases you just take the middle and it'll work.
you don't need to know the value of delta in terms of epsilon at the begenning of the problem.
Just get to the fact that |.....|
I'm confused. Why is it redundant...? You won't be given delta in an actual problem.. He just said, "this is what I came up with," and showed us how he came up with it.
Uh he did go through how he got that delta through the proof though. He just mentioned the value he got beforehand
Thanks its helpful
In the last when you claim |a|/|1+a^2|=1,|a|=1,a
There is A question in my mind as u select epsilon is greater then 0 I agree but by which logic u said that delta will equal to epsilon/2
Um if you decided to set a
Why is abs(a)
at 3:29 why can you just change |Y^2-X^2| to |X^2-Y^2|? Thanks :)
The absolute value of the difference of real numbers is a metric on the real numbers, so you may change the order of the terms and obtain an equivalent expression. More simply, notice that the absolute value of a difference gives the distance on a number line between the two values. This distance shouldn't change just because we change the order of the values.
Ex. 4 - 9 = -5, 9 - 4 = 5, but |-5|=|5| = 5
Thank you!
very helpful, thank you
At 1:49 x,y belongs to A not R.
In this case A = R because our function is from R into R. Thank you for the comment:)
Just did this problem a few days ago, literally did the exact same thing you did, tehe. :)
at 7:14 why can we just simply drop the a^2? btw thank you for this video , helps so much!!
you said that we could just "get rid" of the |1+x^2| and |1+y^2| in denominators but you didn't explain it.
Well, since it is absolute value it must be positive. And of if it is positive it can only makes that whole fraction smaller number. So if you get rid of it it doesnt really matter.
Ivan Ivan It indeed becomes a smaller number, but we claim that is is greater of equal, right? So how can two numbers suddenly be greater when they become smaller?
@@jorthouben denominator became smaller makes the fraction bigger
there is any way to know the function uniformly or not without testing?
Very good example because the use of triangle inequality and even more the second proof. Thank you!
what is the difference between uniformly continuous and continuous?
uniformly continuous is over a set of points while continuous only focuses on one point. This is 3 years too late lmao
+The Math Sorcerer um if memory serves the rule I worked out was to check what is the largest number the 1/(1+x^2)+1/(1+y^2) could be which is 2 in this case the dive epsilon by that number and call it delta.
+ahmad3652 You forgot the absolute values up top, I agree with your statement though yes.
+The Math Sorcerer it's a damn nuisance typing them in but good analysis demands it.
how we apply this claim herw.. how is |x|
I am sooo late I have a doubt...i hope you reply.....i understand the proofs...but is there any tips on determining the delta......i understand it depends on the question but some tips on that
omg youre awesome ! keep up the videos
Fatemeh Moh Thank you! I'm glad it helped someone:)
Thank you so much!
You're saving my ass right now lol thank you!
LOL! np man
how can you just "drop" parts
it's possible in inequalities, he replaces condition with a weaker one
because it's absolute value we know the things he is "dropping" are positive and we also know that the bigger the denominator of a number the smaller the number. So if we're working with inequalities it follows that |x|/ |1+x^2| |1+y^2| < |x|/ |1+x^2|
Thank you sir
👍
Really helpful..
Very happy it helped😃
in your claim after the end of your proof why did you drop the a^2 in |1+a^2| ? And for a>1 why did you drop the 1 in |1+a^2|?
+James Digno a^2>=0 for all a in R, so (1+a^2)>=(1) => mod(1+a^2)>=mod(1) thus dividing by the smaller expression will give a greater expression as it is a reciprocal (laws of inequalities).
A similiar argument will convince you that dropping the 1 will also give the same result.
thanks :)
Great video but please explain how is this function uniformly continuous if when I graph it it does not have a constant slope (isn't that required to be uniformly continuous)
Thx
You are welcome!
What is the logic of choosing $=€/2
Why have a TH-cam channel explaining shit if you dont reply to the comments based on the vids
I still can't see the difference between continuity and uniform continuity. I think I don't understand them correctly. Please please please please please make video on them and show the differences
he has a video on it
I'm your future version. It's both amusing & depressing that you still don't understand them & have to rewatch the video now! Haha
EH MERCEEEEEE FRANKIE
Frannnnnnkounet
You gave the definition of continuity not uniform continuity?
It is uniform because the function is continuous at any accumulation point y. In other words, you're picking any two points and showing continuity.
awesome bro