How to Expand x+1 Raised to an Irrational Power

If n is a natural number, (x+1)^n can be easily expanded, but how can we expand it when it is raised to a general real power?

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[Materials]
VOICEVOX:ずんだもん (Zundamon, illustrated by sakamoto_AHR)
VOICEVOX:四国めたん (Shikoku Metan, illustrated by sakamoto_AHR)
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This video uses synthesized voices.

#math

[ArturRoncatoLopesCarvalho]

imagine a math professor showing this to college students.

[gntompomar]

I never believed that anime girls would teach me math

[christopherjaya]

I expected that gamma function was introduced, but oh well... after all the Γ(x+k)/Γ(x) = x*(x-1)*...*(x-k+1)

[michaelwarnecke]

I'm 5th semester in CS/Math and I never thought about that way to generalize the binomial formula. I vaguely knew about the Gamma function before, but this is such an approachable way to generalize, very well explained.

[fetsexe]

Love the 4th wall break. Nice video, refreshing some math I haven't thought about in a while...

[aleksanderkilinski]

Great idea!
You use anime girls to teach math.
One girl is the teacher and another one is the student.
The best part is that the student try to solve the problem (and make small mistake) and the techer only support her (explain mistake using example).

[mikip]

Wait. Then we can make continuous pascal triangles?

[machine-boy]

This channel is an invaluable source of information providing mathematical intuition on par or better than 3blue1brown

[rexevan]

i Wonder who's target audience here?

[cdkw]

I love that little text at the end of every episode saying you gained this or you unlocked that, it feels like I streaming some math skills and actually getting something

[simran-]

We got anime girls teaching math before GTA 6

[matmagix]

I went to the Taylor expansion of (1+x)^n immediately when I saw this problem but hadn't considered the binomial coefficient for non-natural numbers or where n < k.

[JohnD-qdgo]

This is an unexpected way to present generalized binomial coefficients and expansion. I appreciate that our amine hosts acknowledged that this was an intuitive approach and not a rigorous proof; a proof is probably beyond the scope of an 11-minute video.

[tommasodavi]

I really appreciate the contents and how you tackle them. You made them interesting. Keep doing these videos!
A big thank you

[diogeneslaertius]

My first attempt on this problem would be just to go with the exponential.
E.g. (x+1)^π = e^(π ln (x + 1)) = 1 + π ln (x + 1) + (π ln (x + 1)) ^ 2 / 2 + (π ln (x + 1))^3 / 6 + ...
In this approach we can use complex numbers, there is no limitation for the argument x.
An alternative approach could be in picking an infinite series representation of π and then (x+1)^π becomes an infinite product (x+1)^s1 * (x+1)^s2 * ... where si come from that series representation of π (infinite sum). Now I'm going to watch the video.

[marooo]

genuinely one of the best math channels

[TheDannyAwesome]

Thank you for another great video! But I don't feel you have fully answered the question. Your series expansion only converges for |x| < 1, but the original problem statement does not have this restriction. So further, you should write ( x + 1 ) ^pi = x^pi * ( 1 + 1/x ) ^pi, and the series expansion for this would converge for |x| > 1. Finally supplement with 1^pi = 1 and leave the x = - 1 case for another day.

[noahdonson]

This channel teaches math better than literally any other channel I’ve seen

[Be_Niko]

i love this series so much, my brain is braining now

[AmeMori]

wow I didn't expect this much intuition for the part I was struggling during complex analysis in Mathematics for Theoretical Physics course. Thank you.