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$(1+ta_1)(1+ta_2)...(1+ta_n)=1$ $\forall t\in R$ if and only if $a_1=a_2=...=a_n=0$

Is it true? If yes, how to prove it?

I already gave a proof when $n=3$.

$1=(1+ta_1)(1+ta_2)(1+ta_3)=1+(a_1+a_2+a_3)t+(a_1a_2+a_1a_3+a_2a_3)t^2+(a_1a_2a_3)t^3$

It is equivalent to

$a_1+a_2+a_3=0$

$a_1a_2+a_1a_3+a_2a_3=0$

$a_1a_2a_3=0$

And this gives

$a_1=a_2=a_3=0$

Is it possible to generalize it for arbitrary $n$?

$(1+ta_1)(1+ta_2)\cdots(1+ta_n)=1$, $\forall{t}$ $\in$ $\Bbb{R}$ ⟺ $a_1=a_2=\cdots=a_n=0$

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  • $\begingroup$ Try to use a choose b form for coefficient to generalize to higher n $\endgroup$ Sep 7, 2018 at 2:14
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    $\begingroup$ Instead of messing with formulas, try induction on the number of $a_i$. $\endgroup$ Sep 7, 2018 at 2:21
  • $\begingroup$ Are $\,a_k\,$ assumed to be real or complex? $\endgroup$
    – dxiv
    Sep 7, 2018 at 2:24
  • $\begingroup$ Hint: the corresponding formulas are those which stated in Vieta’s Theorem, I.e. it means that $a_1,a_2,a_3,\ldots ,a_n$ are the roots of the polynomial $x^n=0$. $\endgroup$
    – BAI
    Sep 7, 2018 at 15:10
  • $\begingroup$ A nice exercise! What if the equation is true only for infinite $t$? $\endgroup$
    – dmtri
    Sep 7, 2018 at 15:15

2 Answers 2

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If there exist an $\,a_k \ne 0\,$ then for $\,t = -1/a_k\,$ the LHS is $\,0\,$ i.e. different from the RHS which is $\,1\,$. Therefore the equality can hold for all real $\,t\,$ iff $\,a_k = 0 \mid k=1,2,\ldots,n\,$.

Note: the above assumes that the coefficients $\,a_k\,$ are real.

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  • $\begingroup$ @stressedout Added a note spelling out the underlying assumption, and also asked the OP for clarification. If the $\,a_k\,$ turn out to be complex, I will ammend my answer. $\endgroup$
    – dxiv
    Sep 7, 2018 at 2:28
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    $\begingroup$ I do assume ak are real numbers. Thanks. That's quite a simple way. $\endgroup$ Sep 7, 2018 at 2:31
  • $\begingroup$ What if the numbers were complex? Not sure why it's different (just curious). $\endgroup$ Sep 7, 2018 at 2:33
  • $\begingroup$ @IntegrateThis Then the proof would likely have to go along the line of the other posted answer, but it gets a bit more involved at that point. The simple argument above wouldn't work since $\,t=-1/a_k\,$ would not necessarily be a real number, so the given equality would not be guaranteed to hold. $\endgroup$
    – dxiv
    Sep 7, 2018 at 2:43
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I've just found a new way to prove this.

It's not hard to see that the coefficients of polynomial are all zeros. When $n=4$, that is to say,

$a_1+a_2+a_3+a_4=0$

$a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4=0$

$a_1a_2a_3+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4=0$

$a_1a_2a_3a_4=0$

By the last equality, at least one $a_i(i=1,2,3,4)$ is zero. Let's assume $a_1=0$.

Then by the second last equality, $a_2a_3a_4=0$. Let's assume $a_2=0$.

Then by the third last equality, $a_3a_4=0$. Let's assume $a_3=0$.

Then by the first equality, $a_4=0$.

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  • $\begingroup$ (+1) That works once you fill-in the blanks. That's essentially what (I think) this comment was hinting at. And btw, it works just as well if the $\,a_k\,$ are allowed to be complex. $\endgroup$
    – dxiv
    Sep 7, 2018 at 3:01
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    $\begingroup$ @dxiv: Yep, that was the gist of my comment. The polymomial $(1+ta_1)...(1+ta_n) - 1$ is zero for infinitely many $t$ (all real $t$), so must be identically 0. The coefficient on $t^n$ is the product of the $a_i$, so at least one $a_i$ must be zero. That means your really only had $n-1$ factors, so induction can take over. (And the case case is just the fact that $ta_1 = 0$ for all $t$ implies $a_1 = 0$). $\endgroup$ Sep 7, 2018 at 3:46
  • $\begingroup$ You could use induction to prove your assumption, especially when the equation is supposed to be true for $t$ not in $\Bbb{R}$ but just for infinity many values. $\endgroup$
    – dmtri
    Sep 7, 2018 at 16:05

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