#An interesting (or uninteresting) property I found for polynomials

27 messages · Page 1 of 1 (latest)

thin junco
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For any (a+x)^n, sum of its coefficients is (a+1)^n
sum of its even power coefficients is [(a+1)^n+(a-1)^n]/2
sum of its odd power coefficients is [(a+1)^n-(a-1)^n]/2

Pretty sure somebody else had discovered this milennias ago. Not sure if useful in any way. Ahh

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Yeah Taylor expansion prbly

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but good conclusion to remember

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might come in handy

round onyx
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Prove it or derive it

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If it is not a conjecture or postulate/axiom

thin junco
round onyx
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Then please do so

thin junco
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  1. expand the bionomial to obtain some random ax^n+bx^(n-1)...yx+z
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substitute x with 1 and done

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a+b+c+d+...+y+z=a*1^n+b*1^(n-1)+...+y*1+z

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  1. similarly, substitute x with -1, then add the series from the 1. statement. Odd powers are minus+plus, even powers are plus+plus, so you end with 2 times of even powers, divide by 2
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  1. basically 2. but you substitute with -1 and multiply with -1 on the alternating sequence to invert the parity
thin junco
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(A+1)^n=a+b+c+d+...+y+z
(A-1)^n=a-b+c-d... (assuming n is even)

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smash the 2 together by summing

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2a+2c+2e+2g...

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invert the second line
-(A-1)^n=-a+b-c+d...

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smash

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2b+2d+2f+2h...

rain merlin
# thin junco 2b+2d+2f+2h...

fwiw, yes, your idea on the proof is correct. All 3 of these follow from the binomial theorem (which for all intents and purposes just tells you precisely what the coefficients are)

Though as a general notation comment, since you don't know how many terms there are, don't just use the letters of the alphabet (that implicitly says n=26)

$(A+1)^n=a_0+a_1A+a_2A^2+...+a_nA^n$

late relicBOT
rain merlin
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$(A-1)^n=a_0A^n+a_1A^{n-1}(-1)^1+a_2A^{n-2}(-1)^2+...+a_n(-1)^n$

late relicBOT
thin junco
rain merlin
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Not even the latex, just a general improvement for your argument if you hadn't done that

thin junco
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I thought it'd be messy if I wrote just a0*x^n etc etc