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

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

I hope you are having a great time!

I have presumably found a way to solve

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

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

and \(m\) is any real or complex number.

Claim:

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

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

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

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

Example

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

So \(\left( { r }_{ 1 }+3 \right) \left( { r }_{ 2 }+3 \right) \left( { r }_{ 3 }+3 \right) ={ r }_{ 1 }{ r }_{ 2 }{ r }_{ 3 }+3{ r }_{ 1 }{ r }_{ 2 }+3{ r }_{ 1 }{ r }_{ 3 }+3{ r }_{ 2 }{ r }_{ 3 }+9{ r }_{ 1 }+9{ r }_{ 2 }+9{ r }_{ 3 }+27\)

Then using Vieta's, \({ r }_{ 1 }{ r }_{ 2 }{ r }_{ 3 }+3{ r }_{ 1 }{ r }_{ 2 }+3{ r }_{ 1 }{ r }_{ 3 }+3{ r }_{ 2 }{ r }_{ 3 }+9{ r }_{ 1 }+9{ r }_{ 2 }+9{ r }_{ 3 }+27=-\frac { -120 }{ 2 } +3\times \frac { -14 }{ 2 } -9\times \frac { 12 }{ 2 } +27\\ =12\)

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

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

Proof 1

\(\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right) ={ m }^{ 0 }\prod _{ y=1 }^{ n }{ { r }_{ n } } +{ m }^{ 1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 }{ r }_{ 2 }\dots { r }_{ n-1 } } \right) +\cdots +{ m }^{ n-1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 } } \right) +{ m }^{ n }\left( \sum _{ n=1 }^{ n }{ 1 } \right) \)

Using Vieta's, \({ m }^{ 0 }\prod _{ y=1 }^{ n }{ { r }_{ y } } +{ m }^{ 1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 }{ r }_{ 2 }\dots { r }_{ n-1 } } \right) +\cdots +{ m }^{ n-1 }\left( \sum _{ cyc }^{ }{ { r }_{ 1 } } \right) +{ m }^{ n }\left( \sum _{ n=1 }^{ n }{ 1 } \right) ={ \left( -1 \right) }^{ n }\frac { { m }^{ 0 }{ a }_{ 0 } }{ { a }_{ n } } +\cdots { \left( -1 \right) }^{ 1 }\frac { { m }^{ n-1 }{ a }_{ n-1 } }{ { a }_{ n } } +{ \left( -1 \right) }^{ 0 }\frac { { m }^{ n }{ a }_{ n } }{ { a }_{ n } } \\ =\sum _{ y=0 }^{ n }{ \frac { { \left( -1 \right) }^{ y }{ m }^{ n-y }{ a }_{ n-y } }{ { a }_{ n } } } \\ ={ \left( -1 \right) }^{ n }\frac { f\left( -m \right) }{ { a }_{ n } } \)

Proof 2

With the basic definition of a polynomial,

\({ a }_{ n }{ x }^{ n }+{ a }_{ n-1 }{ x }^{ n-1 }+\dots { a }_{ 1 }{ x }+{ a }_{ 0 }={ a }_{ n }\left( x-{ r }_{ 1 } \right) \left( x-{ r }_{ 2 } \right) \cdots \left( x-{ r }_{ n } \right) \)

\(\left( { r }_{ 1 }+m \right) \left( { r }_{ 2 }+m \right) \dots \left( { r }_{ n }+m \right) ={ \left( -1 \right) }^{ n }\left( -m-{ r }_{ 1 } \right) \left( -m-{ r }_{ 2 } \right) \cdots \left( -m-{ r }_{ n } \right) \\ ={ \left( -1 \right) }^{ n }\frac { f\left( -m \right) }{ { a }_{ n } }\)

Broader definition

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

for \(k\) being any real or complex number.

We too also can solve this by take \(k\) out in each term and then use the generalisation to solve

\(\left( k{ r }_{ 1 }+m \right) \left( k{ r }_{ 2 }+m \right) \dots \left( { kr }_{ n }+m \right) ={ k }^{ n }\left( { r }_{ 1 }+\frac { m }{ k } \right) \left( { r }_{ 2 }+\frac { m }{ k } \right) \cdots \left( { r }_{ n }+\frac { m }{ k } \right) \\ ={ \left( -1 \right) }^{ n }{ k }^{ n }\frac { f\left( -\frac { m }{ k } \right) }{ { a }_{ n } } \\ ={ \left( -k \right) }^{ n }\frac { f\left( -\frac { m }{ k } \right) }{ { a }_{ n } } \)

Phew...

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

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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 · 6 months, 2 weeks ago

@Mehul Arora – thanks! – Joel Yip · 6 months, 2 weeks ago

Good Note. Nice work Joel Yip. – Aditya Sky · 6 months, 1 week ago

Thank you to all of the positive replies to this note! I just want to say thanks again! – Joel Yip · 6 months, 2 weeks ago

nice work Joel. thumbs up!! – Will Jain · 6 months, 2 weeks ago

Should I give more examples? – Joel Yip · 6 months, 2 weeks ago

great note @Joel Yip !!!!! – Shivam Mishra · 6 months, 2 weeks ago

@Shivam Mishra – thank you! – Joel Yip · 6 months, 2 weeks ago

Perhaps you can add questions into this! – Joel Yip · 6 months, 2 weeks ago

@Joel Yip – Try this then

Try this then – Joel Yip · 6 months, 2 weeks ago

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

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