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If $\alpha=e^{\frac{i2\pi}{7}}$ and $f(x)= a_0+\sum_{k=1}^{20} a_kx^k$ then the value of $$f(x)+f(\alpha x)+f(\alpha^2 x)+\cdots+f(\alpha^6 x)$$ is $ka_0$. Then find the value of $k$.

I used a common property of complex numbers which is giving me entire different result. It is as follows. $$1^k+\alpha^k+(\alpha^2)^k+\cdots+(\alpha^{n-1})^k=0$$ if $k$ in not a multiple of $n$, else $n$; where $\alpha$ is the $n^{th}$ root of unity.

Now from the above fact if can be said that there are some powers of $x$ which are multiple of $7$, hence the required sum will contain coefficients other than $a_0$. Hence I highly doubt the question to be misprinted in my book. Please help whether I am right or wrong.

Michael Hardy
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sameed hussain
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2 Answers2

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I think a silly confusion is caused by the use of bad notations in the question. The $k$ in the definition of $f$ is a dummy variable which has nothing to do with the $k$ in the statement that the sum is $ka_0$. Change $k$ to $i$ or some other variable in the definition of $f(x)$. Note that $(1-\alpha)(1+\alpha+\alpha^{2}+\cdots+\alpha^{6})=1-\alpha^{7}=0$. So $1+\alpha+\alpha^{2}+\cdots+\alpha^{6}=0$ Now if you compute the sum you wll get $7a_0$ so $k=7$.

Kavi Rama Murthy
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    I am confused by your answer. I independently came to the exact same conclusion as the OP. That is, when you compute the 20-th degree polynomial, the coefficients of each term are zero, except for the coefficients that pertain to the terms $x^0, x^7$ and $x^{14}$. For those terms, I agree with the OP that the coefficients are each $(7)$. So you have that $7a_0 + 7a_7x^7 + 7a_{14}x^{14} = ka_0$. I don't see how that you deduce from this that $k = 7$, since that inference would require that $(a_7x^7 + a_{14}x^{14}) = 0,$ for any value of $(x).$ – user2661923 Apr 24 '21 at 13:49
  • One interpretation of the problem that would resolve the issue is that since you are given that the final sum equals $ka_0$, for any value of $(x)$, you are forced to conclude that $a_7 = 0$ and $a_{14} = 0$. A somewhat convoluted inference, but still feasible. Is this what you are thinking? – user2661923 Apr 24 '21 at 13:53
  • @user2661923 Please read my answer now. – Kavi Rama Murthy Apr 24 '21 at 23:31
  • I just left an answer in response. In short, I agree that the variable $k$ is overloaded in the original presentation of the question, but still think that the convoluted inference is necessary. – user2661923 Apr 25 '21 at 03:14
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This answer is an extension of the comments that I left after Kavi Rama Murthy's answer, and explains why I agree with the OP's analyis that the problem is puzzling. I also explain why the problem requires the convoluted inference that both $a_7$ and $a_{14}$ are equal to zero.

First of all, I definitely agree with almost all of the analysis in Kavi Rama Murthy's answer:

  • $1 + \alpha + \alpha^2 + \cdots \alpha^6 = 0.$

  • In the original presentation of the problem, the variable $k$ is overloaded. This is easily resolved by re-expressing $f(x)$ as
    $\displaystyle a_0 + \sum_{r=1}^{20} a_r x^r.$

  • For $r \in \Bbb{Z^+}$ such that $r$ is not a multiple of $(7)$
    $\displaystyle 1 + (\alpha^r) + (\alpha^r)^2 + (\alpha^r)^3 + (\alpha^r)^4 + (\alpha^r)^5 + (\alpha^r)^6 = 0.$
    This is because Kavi Rama Murthy's analysis against $(\alpha)$ also pertains to $(\alpha)^r.$
    Namely that $(1 - [\alpha^r]) \neq 0$, while
    $\displaystyle (1 - [\alpha^r]) \times (1 + [\alpha^r] + [\alpha^r]^2 + \cdots + [\alpha^r]^6) ~=~ 1 - [\alpha^r]^7 = 0.$

However, as explained below, this does not resolve the conflict originally identified by the original poster.

Let $\displaystyle g(x) = f(x) + f(\alpha x) + f(\alpha^2 x) + \cdots + f(\alpha^6 x)$

$\displaystyle =~ a_0 + \sum_{r=1}^{20} a_r x^r $

$\displaystyle +~ a_0 + \sum_{r=1}^{20} a_r (\alpha^r) x^r $

$\displaystyle +~ a_0 + \sum_{r=1}^{20} a_r (\alpha^{2r}) x^r $

$\displaystyle +~ a_0 + \sum_{r=1}^{20} a_r (\alpha^{3r}) x^r $

$\displaystyle +~ a_0 + \sum_{r=1}^{20} a_r (\alpha^{4r}) x^r $

$\displaystyle +~ a_0 + \sum_{r=1}^{20} a_r (\alpha^{5r}) x^r $

$\displaystyle +~ a_0 + \sum_{r=1}^{20} a_r (\alpha^{6r}) x^r. $

Therefore,

$$g(x) = 7a_0 + \sum_{r=1}^{20} \left[a_r x^r \left(\sum_{s=0}^6 \alpha^{(rs)} \right) \right]. \tag{1}$$

As discussed, as $r$ takes on the values $(1)$ through $(20)$,
if $r$ is not a multiple of $(7)$, then
the inner summation for $g(x)$ in equation (1) above,
$\displaystyle \left(\sum_{s=0}^6 \alpha^{(rs)} \right)$
will equal zero.

However, for $r = 7$ or $r = 14$
the inner summation for $g(x)$ in equation (1) above,
$\displaystyle \left(\sum_{s=0}^6 \alpha^{(rs)} \right)$
will instead equal $(7).$

Therefore

$$g(x) = 7a_0 + 7a_7(x^7) + 7a_{14}(x^{14}). \tag{2}$$

A premise is given that (presumably) for all values of $x, ~g(x) = ka_0$. I don't see how this premise can be true, for all values of $x$, unless both $a_7$ and $a_{14}$ are equal to zero.

Therefore, since the constraint that for all values of $x, ~g(x) = ka_0$ is a premise,
one is forced into the convoluted inference that $0 = a_7$ and $0 = a_{14}$.

user2661923
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