Generalisation of (r1+m)(r2+m)(rn+m)\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right)

Hello every 45π4096(72.5)\frac { 45\pi }{ 4096 } \left( \begin{matrix} 7 \\ 2.5 \end{matrix} \right) .

I hope you are having a great time!

I have presumably found a way to solve

(r1+m)(r2+m)(rn+m)\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right)

if rn{ r }_{ n } are the roots to any equation of the form f(x)=anxn+an1xn1+a1x+a0f\left( x \right)={ a }_{ n }{ x }^{ n }+{ a }_{ n-1 }{ x }^{ n-1 }+\dots { a }_{ 1 }{ x }+{ a }_{ 0 }

and mm is any real or complex number.

Claim:

(r1+m)(r2+m)(rn+m)=(1)nf(m)an\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right) ={ \left( -1 \right) }^{ n }\frac { f\left( -m \right) }{ { a }_{ n } }

if f(x)=anxn+an1xn1+a1x+a0f\left( x \right)={ a }_{ n }{ x }^{ n }+{ a }_{ n-1 }{ x }^{ n-1 }+\dots { a }_{ 1 }{ x }+{ a }_{ 0 }

Simply put an{ a }_{ n } is the coefficient of xn{ x }^{ n } in f(x)f\left( x \right) or even simpler,

an{ a }_{ n } is the leading coefficient of f(x)f\left( x \right).

Example

Lets have an explicit example where m=3m=3 and f(x)=2x3+12x214x120f\left( x \right) =2{ x }^{ 3 }+12{ x }^{ 2 }-14x-120

So (r1+3)(r2+3)(r3+3)=r1r2r3+3r1r2+3r1r3+3r2r3+9r1+9r2+9r3+27\left( { r }_{ 1 }+3 \right) \left( { r }_{ 2 }+3 \right) \left( { r }_{ 3 }+3 \right) ={ r }_{ 1 }{ r }_{ 2 }{ r }_{ 3 }+3{ r }_{ 1 }{ r }_{ 2 }+3{ r }_{ 1 }{ r }_{ 3 }+3{ r }_{ 2 }{ r }_{ 3 }+9{ r }_{ 1 }+9{ r }_{ 2 }+9{ r }_{ 3 }+27

Then using Vieta's, r1r2r3+3r1r2+3r1r3+3r2r3+9r1+9r2+9r3+27=1202+3×1429×122+27=12{ r }_{ 1 }{ r }_{ 2 }{ r }_{ 3 }+3{ r }_{ 1 }{ r }_{ 2 }+3{ r }_{ 1 }{ r }_{ 3 }+3{ r }_{ 2 }{ r }_{ 3 }+9{ r }_{ 1 }+9{ r }_{ 2 }+9{ r }_{ 3 }+27=-\frac { -120 }{ 2 } +3\times \frac { -14 }{ 2 } -9\times \frac { 12 }{ 2 } +27\\ =12

and indeed (1)3f(3)2=12{ \left( -1 \right) }^{ 3 }\frac { f\left( -3 \right) }{ 2 } =12

and also the roots of f(x)=2x3+12x214x120f\left( x \right) =2{ x }^{ 3 }+12{ x }^{ 2 }-14x-120 is 5-5,4-4 and 33. Meaning the product is (5+3)(4+3)(3+3)=12\left( -5+3 \right) \left( -4+3 \right) \left( 3+3 \right) =12

Proof 1

(r1+m)(r2+m)(rn+m)=m0y=1nrn+m1(cycr1r2rn1)++mn1(cycr1)+mn(n=1n1)\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right) ={ m }^{ 0 }\prod _{ y=1 }^{ n }{ { r }_{ n } } +{ m }^{ 1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 }{ r }_{ 2 }\dots { r }_{ n-1 } } \right) +\cdots +{ m }^{ n-1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 } } \right) +{ m }^{ n }\left( \sum _{ n=1 }^{ n }{ 1 } \right)

Using Vieta's, m0y=1nry+m1(cycr1r2rn1)++mn1(cycr1)+mn(n=1n1)=(1)nm0a0an+(1)1mn1an1an+(1)0mnanan=y=0n(1)ymnyanyan=(1)nf(m)an{ m }^{ 0 }\prod _{ y=1 }^{ n }{ { r }_{ y } } +{ m }^{ 1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 }{ r }_{ 2 }\dots { r }_{ n-1 } } \right) +\cdots +{ m }^{ n-1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 } } \right) +{ m }^{ n }\left( \sum _{ n=1 }^{ n }{ 1 } \right) ={ \left( -1 \right) }^{ n }\frac { { m }^{ 0 }{ a }_{ 0 } }{ { a }_{ n } } +\cdots { \left( -1 \right) }^{ 1 }\frac { { m }^{ n-1 }{ a }_{ n-1 } }{ { a }_{ n } } +{ \left( -1 \right) }^{ 0 }\frac { { m }^{ n }{ a }_{ n } }{ { a }_{ n } } \\ =\sum _{ y=0 }^{ n }{ \frac { { \left( -1 \right) }^{ y }{ m }^{ n-y }{ a }_{ n-y } }{ { a }_{ n } } } \\ ={ \left( -1 \right) }^{ n }\frac { f\left( -m \right) }{ { a }_{ n } }

Proof 2

With the basic definition of a polynomial,

anxn+an1xn1+a1x+a0=an(xr1)(xr2)(xrn){ a }_{ n }{ x }^{ n }+{ a }_{ n-1 }{ x }^{ n-1 }+\dots { a }_{ 1 }{ x }+{ a }_{ 0 }={ a }_{ n }\left( x-{ r }_{ 1 } \right) \left( x-{ r }_{ 2 } \right) \cdots \left( x-{ r }_{ n } \right)

(r1+m)(r2+m)(rn+m)=(1)n(mr1)(mr2)(mrn)=(1)nf(m)an\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right) ={ \left( -1 \right) }^{ n }\left( -m-{ r }_{ 1 } \right) \left( -m-{ r }_{ 2 } \right) \cdots \left( -m-{ r }_{ n } \right) \\ ={ \left( -1 \right) }^{ n }\frac { f\left( -m \right) }{ { a }_{ n } }

Broader definition

(kr1+m)(kr2+m)(krn+m)\left( k{ r }_{ 1 }+m \right) \left( k{ r }_{ 2 }+m \right) \dots \left( { kr }_{ n }+m \right)

for kk being any real or complex number.

We too also can solve this by take kk out in each term and then use the generalisation to solve

(kr1+m)(kr2+m)(krn+m)=kn(r1+mk)(r2+mk)(rn+mk)=(1)nknf(mk)an=(k)nf(mk)an\left( k{ r }_{ 1 }+m \right) \left( k{ r }_{ 2 }+m \right) \dots \left( { kr }_{ n }+m \right) ={ k }^{ n }\left( { r }_{ 1 }+\frac { m }{ k } \right) \left( { r }_{ 2 }+\frac { m }{ k } \right) \cdots \left( { r }_{ n }+\frac { m }{ k } \right) \\ ={ \left( -1 \right) }^{ n }{ k }^{ n }\frac { f\left( -\frac { m }{ k } \right) }{ { a }_{ n } } \\ ={ \left( -k \right) }^{ n }\frac { f\left( -\frac { m }{ k } \right) }{ { a }_{ n } }

Phew...

I really do hope this note helped you in one way or another! Well, that's probably it!

Note by Joel Yip
3 years, 8 months ago

No vote yet
1 vote

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Comments

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Nice note! @Joel Yip

Yes, I think this is a good generalization for such sums. No wonder we'll get to see more good problems from many others now :)

Mehul Arora - 3 years, 8 months ago

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thanks!

Joel Yip - 3 years, 8 months ago

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Perhaps you can add questions into this!

Joel Yip - 3 years, 8 months ago

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Try this then

Try this then

Joel Yip - 3 years, 8 months ago

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great note @Joel Yip !!!!!

shivam mishra - 3 years, 8 months ago

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thank you!

Joel Yip - 3 years, 8 months ago

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Should I give more examples?

Joel Yip - 3 years, 8 months ago

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nice work Joel. thumbs up!!

will jain - 3 years, 8 months ago

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Thank you to all of the positive replies to this note! I just want to say thanks again!

Joel Yip - 3 years, 8 months ago

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Good Note. Nice work Joel Yip.

Aditya Sky - 3 years, 7 months ago

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