[Discrete Mathematics] Coefficient Extraction and Extended Binomial Theorem

Today we look at the extended binomial coefficient and ways to extract the coefficients of generating functions.

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[MrEyee]

Any one watching this, please note that the answer to the last exercise is "19 choose 14" as others have noted.
The author made a slight mistake while factoring out x^6 from (x^6 + x^7 + x^8 + ... )^6

The right way is as follows:
(x^6 + x^7 + x^8 + ...)^6
Factor out x^6, from the series in the brackets, everything is still to power 6.
= (x^6(1 + x + x^2 + ... ))^6
By the rules of algebra, the outside exponent 6 goes into each term in the big bracket
= x^36(1 + x + x^2 + ...)^6
Then substitute the power series in the bracket with its closed form "Don't forget to raise it to power 6" -- that was his mistake.

When we arrived at different answers, I spent some time figuring out where I had made a mistake, but later discovered I was correct. Hope this helped someone.

[mynameisisu]

In the final question the answer should be 19c14. You did a mistake the forgot to write the power of 6 in going from step 1 to step 2. :)

[TheAurians]

Just want to say, these are really great videos. These are really helping me to understand the core concepts and applications rather than just staring blankly at in class.

[genssi]

Wow, Your video is a great explanation of K.Rosen's book on generating functions!Thank you!

[Alex-fhmy]

Trev, you are a hero. Better yet, you are MY hero.

[ZelenoJabko]

The final task is not 1, but 19 choose 14.

[jessstuart]

10:00 you might mention it is important to select the binomial term (a+b) so that you get x^i power in the binomial expansion, where i is the exponent of the x-coefficient you are trying to find. Choose a and b wrong, and you end up with a x^(n-i) term instead, which is not what you want.

Numerical values...
[x^18] = 19188950
[x^50]= C(19, 5) = C(19, 14) = 11628

[pokoknyaakuimut]

Wow that is so good content. I don't think the easier way to solve my difficulty like that previously. Thank you very much 🙏🏻🙏🏻🙏🏻

[comuniunecuosho-campulbudi]

the best math channel is newcalculus of John Gabriel

[didles]

I think that:
(x^6+x^7+x^8+...)^6 = ((x^6)(1+x+x^2+...))^6 = x^36 (1-x)^(-6) 
But you dropped that exponent for some reason.

[gloriayoon]

Thank you for this. It's hard to explain just how helpful this was for macm :)

[marcushendriksen]

Goddamn, I really love generating functions!!! They're amazingly powerful

[mriamvanstapele]

x^14 -> 1^6(1-x)^-6 -> -6 choose 14 = 19 choose 14?

[tamarshai]

Hi, how do you ectract coefficient from this function:
(x+x²+x⁴+x⁵+x⁷+x⁸)²?
General way to extract from similar functions?

[tkavranoglu]

Its been very helpful for me to understand this subject. Thank you for that but i didn't accept one thing in the video. Please correct me if i am wrong.
in 23.03 i think you have made mistake and as a result you have found solution incorrect. (1/1-x) has to be (1/1-x)^6 and it will be end up as result C(-6, 14) isn't it?

[VedanttMehtaa]

In the end of the question. It should be 1/1-x raise to power 6. Isnt it???

[davemartin]

In the final question, the answer is 1 because he made a mistake when he wrote ^6, when he intended ^1.

[matniet]

Can this be generalized for when I have more than two factors?

[quinndougherty]

15:50 i can't figure out why [30-12] makes 12-12 on x but not 12-12 on 1/(1-x).
Then around 22:40 we multiply 6*6 on the front part which amounts to 6-6 on the main 1/(1-x) part. It seems like the problems draw opposing conclusions.

[ujjwalpawar]

Hey man can you give any source for more practice question.