Changing Order of a Double Summation

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blackpenredpen
100/(1-x)

[brandondrumheller]

"This is not an integral sign so don't get too excited." 😄

[blackpenredpen]

[from my latest post] Hi all,




Lars, Peyam and I are planning an livestream interview. So please leave a question that you'd like to know about Lars. Also if you have a good way to do a livestream interview, please let me know as well. Thank you!


Backstory




I postponed the "Lar's Integral" but my 100 integrals video was born. I really want to thank everyone who had left a supporting comment that help to put a smile on Lar's face during the tough time.


*LARS IS A STAGE 3 CANCER SURVIVOR NOW AND I AM BEYOND HAPPY*


Thank you all, and I wish everyone have a great great life!
100/(1-x)

[AndrewShepherd-nm]

This concept has been crushing me. Great vid. Seeing the inequalities made it clear.

[vellyxenya]

Thank you very much blackpenredpen, you are the first one giving me a true intuition on why the inversion of order of summation is done like that. It's been bugging for a while and I feel freed now! Thank you for this awesome video and all of your other content!

[whyit]

Great video! I almost thought that there _was_ an integral at 5:58 and I got suprised... Congratulations on 300, 000 subscribers! I'm glad to be one of them.That was a very interesting concept, and I can't wait for the next video!

[davidwright]

I knew it wasn't an integral sign but I still got excited, because I've been banging my head against this for too long! I tried this among other ways, which (I now know) were wrong.
Now I know which is right, and why - hence the excitement! Excellent!

[postbodzapism]

Or you can draw a table

m 1 2 3 4 ...
n
1 x
2 x x
3 x x x
4 x x x x

.
.
.
changing horizontal summation to vertical summation will get you the upper and lower limits of summation.

[konstantinoskompothekras]

Thank you friend!
Helped me a lot with a formula with probability generating functions where I couldn't understand why indices change when you change the order of summation

[nicolasgoubin]

"Let me tell you that the answer for this :




:)

[MrHK]

It might be easier to do this way: First take 1/2^m out side and inside is geometric summation so then you are left with two geometric series and you get 1-sum(1/4^m) and you get 1-1/3=2/3 so YAY

[MathForLife]

Congrats with 300k!! I love double summation! Nice video!:)

[legendarycosmo]

Hey blackpenredpen, could you do a video on divergent series?
I really seem to be having a trouble in grasping the concept.

[ChepemnsTr]

Without changing the order of the double summation, first take the 1/2^m out of the inner summation since the index variable that changes is n, then use the fact that the inner summation is a finite geometric series with common ratio r = 1/2 to express it in terms of m (there's a formula for this), and finally evaluate the infinite geometric series.

[amirrezajamshidi]

Finally an easier explanation.Thank you

[megatron]

Great video. This explanation was the best !!!! 😃

[aashsyed]

3:23 what if m was starting at 2? what would we do?

[BobChao]

I did this... really poorly, but was pleased when I also got the correct answer! I expanded out a handful of terms from the series and looked for a pattern. I found that each exponent of 2 was repeated floor(exponent/2) times. So 2^-2 and 2^-3 showed up once, 2^-4 and 2^-5 showed up twice, 2^-6 and 2^-7 showed up thrice, etc. This let me change from a double sum into a single sum: Sum of i=2 -> Infinity of 2^-i * floor(i/2), but that's not something I could work with nicely. So I expanded out sample terms from that and started simplifying, getting an easy 2^-1+2^-3+2^-5+2^-8, oh, so close. Expanding more terms got me the 2^-7 and a 2^-10 with a long tail of numbers ready to collapse it. Good enough for me to suspect my pattern was close, so I wrote down Sum of j=0 -> Infinity of 2^(2j+1) and then solved that infinite geometric series.

[Dragonborn]

@blackpenredpen
It seems like this problem has a sort of symmetry with m and n. What if it didn't, specifically what if it was originally 1/2^(m-n)?

[VibingMath]

Love the magic sound effect 😁
Good video on double summation 👍

[eckhardtdom]

2:02, can someone please explain this step again? Why des m=n there?