In the second case (8:00) partition P1 and P2 create equal sums, because both of them contain the set of points where the step function is discontinuous. Couldn't P3 then also be chosen to be the intersection of P1 and P2 to lead to the same conclusion? Edit: Oh, I just watched the next episode. Now I see the benefit of using the union.
Good Job I mean... maths is my life Now English language could be my life too because I understand math with this language and with your explanation. I am from North of Africa.
Why is the area under the graph the same for a finer partition? If we have no partition at all for example, and took the value of the rectangle with the first constant value, then we added ik more partitions at the jump points where it jumps to lower values, wouldn't the area get smaller? Doesn't this only work if the jump points are always included?
@@brightsideofmaths damn thanks for the quick answer that is pretty impressive. Unfortunately I still don't get it. Isn't the point that the area is the same ragardeless of the partition? It seems to only work when the value of the function doesn't change within each partial segment, otherwise which value would you choose to multiply the segment length with?
In the last part of explaining "Case 2" we have that Sigma_p1 = Sigma_p3 and Sigma_p2 = Sigma_p3. It is visible that no matter how many elements do P1 and P2 have and what are the lengths of each segment, all partitions P1, P2 and P3 cover the whole x-axis. However the c_j values (for P1) and d_j values (for P2) are not the same. This is therefore a bit unclear to me to see how the total sum of area under the two partitions turn out to be the same (the same as the total area under P3) !
I guess now I can see the point. No matter in which way we split the compact set [a,b] in the process of partitioning, the values of step-function phi is not going to change over certain intervals (the values c_j are not changing). and therefore P3 = P1 U P2 is only split the interval [a,b] into more segments. This is why the area-under-phi using P1,P2 or P3 remains the same.
For Case 1: if in P2 for example x2 tilde is involved, left from x3 tilde and therefore left from x1 in P1, then the area of P2 is greater then P1, isn't it? And the condition P1 is a subset of P2 still holds.Edit: nevernimd, x0 = a is allways included, in both sets.
I think it may have been more helpful to define the integral in terms of Riemann sums first. Then it would have been clear why exactly the integral is well-defined in these cases.
Could you be perhaps from the North African country, Tunisia, or any other Maghrebi country that uses French as the main language in the educational system?
![Real Analysis 59 | Integration by Partial Fraction Decomposition [dark version]](/img/n.gif)
I didn't understand this course in French or Arabic my first language! & Now i understand it from you in English 😂 i'm so happy ❤
Thanks :D
Exactly the same 😂
Me too
This is extremely helpful for me to teach my students about the Riemann integral.
Thank you very much :) I hope you can also the other videos for your students :)
What an excellent Christmas present!
Great series on Real Analysis!
In the second case (8:00) partition P1 and P2 create equal sums, because both of them contain the set of points where the step function is discontinuous. Couldn't P3 then also be chosen to be the intersection of P1 and P2 to lead to the same conclusion?
Edit: Oh, I just watched the next episode. Now I see the benefit of using the union.
You are amazing sir, a real life sever thank you very much I now have a hope for my CA😊
Thanks! What is a CA?
Good Job
I mean... maths is my life
Now English language could be my life too because I understand math with this language and with your explanation.
I am from North of Africa.
Thanks! I hope that the subtitles help a little bit :)
Why is the area under the graph the same for a finer partition? If we have no partition at all for example, and took the value of the rectangle with the first constant value, then we added ik more partitions at the jump points where it jumps to lower values, wouldn't the area get smaller? Doesn't this only work if the jump points are always included?
Thanks for the question. Even with more points at the lower level, you would not reduce the size of the rectangle at the higher level :)
@@brightsideofmaths damn thanks for the quick answer that is pretty impressive.
Unfortunately I still don't get it. Isn't the point that the area is the same ragardeless of the partition? It seems to only work when the value of the function doesn't change within each partial segment, otherwise which value would you choose to multiply the segment length with?
Maybe just misunderstood the definition of the function phi? It's already defined as a step function with a chosen partition.@@awesomecraftstudio
For the second case: P1 and P2 must still contain the jumps, right? How do you denote that?
In the last part of explaining "Case 2" we have that Sigma_p1 = Sigma_p3 and Sigma_p2 = Sigma_p3. It is visible that no matter how many elements do P1 and P2 have and what are the lengths of each segment, all partitions P1, P2 and P3 cover the whole x-axis. However the c_j values (for P1) and d_j values (for P2) are not the same. This is therefore a bit unclear to me to see how the total sum of area under the two partitions turn out to be the same (the same as the total area under P3) !
I guess now I can see the point. No matter in which way we split the compact set [a,b] in the process of partitioning, the values of step-function phi is not going to change over certain intervals (the values c_j are not changing). and therefore P3 = P1 U P2 is only split the interval [a,b] into more segments. This is why the area-under-phi using P1,P2 or P3 remains the same.
For Case 1: if in P2 for example x2 tilde is involved, left from x3 tilde and therefore left from x1 in P1, then the area of P2 is greater then P1, isn't it? And the condition P1 is a subset of P2 still holds.Edit: nevernimd, x0 = a is allways included, in both sets.
What is that symbol really called? I mean that circle with a verticle stroke in it.
Is it a step function?
It is phi, a lowercase greek letter :)
Shouldn't the union of the two partitions cover both the cases?
I think it may have been more helpful to define the integral in terms of Riemann sums first. Then it would have been clear why exactly the integral is well-defined in these cases.
Indeed that is possible but I like the step function approach :)
Examples? How to calculate ci?
What do you mean exactly?
Nice :D
I didn't understand this course in French or Arabic my first language! & Now i understand it from you in English 😂 i'm so happy ❤
Thanks :)
Could you be perhaps from the North African country, Tunisia, or any other Maghrebi country that uses French as the main language in the educational system?