An awesome generalisation- of Bashing Unavailable

If you were given the three equations \[\color{Blue}{a+b+c=1}\]\[\color{Blue}{a^2+b^2+c^2=2}\] \[\color{Blue}{a^3+b^3+c^3=3}\]

And then the values of an+bn+cn\color{#D61F06}{a^n+b^n+c^n} are asked for some nZ+n \in \mathbb{Z}^+, then you can get the answer by long expansions, squaring, getting needed terms etc, but there is a smart way too !

Some days ago, i had shared a set by name Bashing Unavailable...Awesome problems , which was on problems of this type , for n{4,5,6,7,8,9,10}n \in \{ 4,5,6,7,8,9,10\}.

If you haven't tried the set, I prefer telling you to try it before you read this note.

You can get the thing for n=4n=4 by algebraic manipulations (and also for further values of nn ) , for example this "part of the solution" which I knowingly wrote to the first problem of the set.

img img

This way, you can get the values.... but here goes the smarter way, the generalisation\color{#69047E}{\textbf{generalisation}}. My way is similar to my friend Aamir Faisal Ansari,a worth following person's way...

See that as shown in the image, we can get values of ab+bc+ca=12ab+bc+ca=\dfrac{-1}{2} and abc=16abc=\dfrac{1}{6}

Next, let's define a sequence {tn}\{t_n\} as tn=an+bn+cnt_n=a^n+b^n+c^n

Then, we do the following algebraic manipulation which will give us the generalisation

an+bn+cn=(a+b+c)(an1+bn1+cn1)<something>a^n+b^n+c^n=(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1}) - \text{<something>}

(something = the extra terms that will come in the expansion of first term of the RHS)

an+bn+cn=(a+b+c)(an1+bn1+cn1)(acn1+bcn1+abn1+cbn1+ban1+can1)a^n+b^n+c^n=(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1}) \\ \quad \quad \quad \quad -(ac^{n-1} + bc^{n-1}+ab^{n-1}+cb^{n-1}+ba^{n-1}+ca^{n-1})

For the extra terms, we have

(acn1+bcn1+abn1+cbn1+ban1+can1)=(ab+bc+ca)(an2+bn2+cn2)<other something>(ac^{n-1} + bc^{n-1}+ab^{n-1}+cb^{n-1}+ba^{n-1}+ca^{n-1}) = \\ \quad \quad \quad \quad \quad \quad (ab+bc+ca)(a^{n-2}+b^{n-2}+c^{n-2}) - \text{<other something>}

Other something=bcan2+cabn2+abcn2=abc(an3+bn3+cn3)\textbf{Other something}= bca^{n-2}+cab^{n-2}+abc^{n-2} = abc(a^{n-3}+b^{n-3}+c^{n-3})

Thus, we have

an+bn+cn=(a+b+c)(an1+bn1+cn1)(ab+bc+ca)(an2+bn2+cn2)+abc(an3+bn3+cn3)a^n+b^n+c^n=(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1})\\\quad \quad \quad\quad \quad - (ab+bc+ca)(a^{n-2}+b^{n-2}+c^{n-2}) \\ \quad \quad \quad \quad\quad+ abc(a^{n-3}+b^{n-3}+c^{n-3})

From this we get the recurrence relation tn=(a+b+c)tn1(ab+bc+ca)tn2+(abc)tn3t_n=(a+b+c) t_{n-1} -(ab+bc+ca)t_{n-2} +(abc)t_{n-3}

And because we know the values of all the coefficients in this thing, we get the recurrence relation tn=tn1+12tn2+16tn3t_n=t_{n-1} +\frac{1}{2} t_{n-2}+\frac{1}{6} t_{n-3}

From this recurrence relation, because you already know the first 3 terms\color{#20A900}{\text{because you already know the first 3 terms}}, you can get the value of all the further terms like a cakewalk !

Reshare if this is helpful, and try to solve the last part of it, the 6th6^{th} part of the set - Bashing Unavailable Part 6

Note by Aditya Raut
6 years, 11 months ago

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This method is known as "Newton's Sums".

Daniel Liu - 6 years, 11 months ago

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@Aditya Raut What about "Bashing Unbelievable Part 1.5"?

Satvik Golechha - 6 years, 11 months ago

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It will come after I release Bashing Unavailable part k=217(2k+k2)0sin2014x dxlog9.7888\displaystyle \color{#69047E}{\textbf{Bashing Unavailable part }} \color{#20A900}{\sqrt{\dfrac{\sum_{k=\sqrt{2}} ^{\sqrt{17}} \bigl( 2^{\sqrt{k}} + k^{\sqrt{2}} \bigr)}{\int_0 ^\infty \sin^{2014} x \text{ dx} - \log_{9.7} 888 }}}

Any more doubts @Satvik Golechha ???

Aditya Raut - 6 years, 11 months ago

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Satvik Golechha - 6 years, 11 months ago

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It doesn't converge according to Wolfram Alpha

Abdur Rehman Zahid - 6 years, 4 months ago

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@Abdur Rehman Zahid Oh man! That was just a troll, because I was meaning to tell Satvik there's no 1.5 coming up, this thing was just for integers... Anyway, thanks for telling, but that big term is not seriously typed !

Aditya Raut - 6 years, 4 months ago

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Thanks a lot!!!

Aamir Faisal Ansari - 6 years, 11 months ago

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Similar technics were given in Arthur Engel , i read them little while ago. BTW Nice work Aditya!!

Sanjeet Raria - 6 years, 10 months ago

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Thanks a lott!! @Aditya Raut !! A wonderful Note!! i have been struggling at questions of these type a bit...but not anymore...thanks!! :):)

A Former Brilliant Member - 6 years, 10 months ago

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Really amazing one...

Sriram Vudayagiri - 6 years, 5 months ago

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What's amazing about the set is that even though you have so many equations in the same three variables, you can't find their values. I mean, I know you're transforming the given equations to the new ones, so you technically can't find the values of the variables, but still.

Omkar Kulkarni - 6 years, 6 months ago

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this is cool bro.!! :)

Satyabrata Dash - 5 years, 1 month ago

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The Newton Sums in 3 variables are:

p=a+b+cq=ab+bc+acr=abcp=a+b+c \\ q=ab+bc+ac \\ r=abc

These are fairly straightforward to calculate if you're give a,a2,a3\sum a,\sum a^2,\sum a^3.

If you let a,b,ca,b,c be the roots of the monic polynomial f(x)f(x) then you have:

f(x)=(xa)(xb)(xc)=x3px2+qxrf(a)=f(b)=f(c)=0f(a)=0    a3pa2+qar=0    a3=pa2qa+rf(x)=(x-a)(x-b)(x-c)=x^3-px^2+qx-r \quad \quad f(a)=f(b)=f(c)=0 \\ f(a)=0 \implies a^3-pa^2+qa-r=0 \implies a^3=pa^2-qa+r

If we let tn=ant_n=\sum a^n then we get:

a3=pa2qa+r    an+3=pan+2qan+1+ranan+3=pan+2qan+1+rantn+3=p  tn+2q  tn+1+r  tna^3=pa^2-qa+r \implies a^{n+3}=p a^{n+2}-q a^{n+1}+r a^n \\ \sum a^{n+3}=p \sum a^{n+2} -q \sum a^{n+1}+r \sum a^n \\ t_{n+3}=p \; t_{n+2}-q \; t_{n+1}+r \; t_n

This gives you a way of quickly calculating tnt_n for large nn as you will know p,q,rp,q,r.

I've used this method on this problem.

Sam Bealing - 5 years, 1 month ago

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Satyabrata Dash - 5 years, 1 month ago

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