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I want to find a linear map $f$ with the property $f(x^n)=nf(x)$.

Suppose we have the recurrence $x_n+\frac{1}{x_n}=x_{n+1}$ with $x_1=a$. If such a map exists then $0=f(x_{n+1})$. So $f$ attains roots at $x_2,x_3,....$.

Does $\{x_2,x_3,....\}$ account for all real numbers if we are allowed to vary $a \in \mathbb{R}$?

Ahmed S. Attaalla
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    By linear function, do you mean $f(x) = ax + b$? – Ennar Jan 01 '17 at 00:48
  • No $f(c_1x+c_2y)=c_1f(x)+c_2(y)$ @Ennar – Ahmed S. Attaalla Jan 01 '17 at 00:49
  • There is no way ${x_2,x_3,\ldots}$ contains all real numbers since ${x_2,x_3,\ldots}$ is a countable set. – JimmyK4542 Jan 01 '17 at 00:51
  • In any case, fix $n$. If $f\neq 0$, you get a polynomial of degree $n$, which can't have more than $n$ roots, contradiction. – Ennar Jan 01 '17 at 00:52
  • don't call it a linear map , by doing so you're saying that you can speak about the power of a vector which is immpossible , just call it a morphisme insteas of a linear map – Seginus Jan 01 '17 at 00:58
  • @Seginus, no, that is untrue, you can take powers of vectors in any algebra over a field. – Ennar Jan 01 '17 at 01:02
  • I've never heard about that , what your definition of a power of a vector ? – Seginus Jan 01 '17 at 01:09
  • @Seginus, please see Algebra over a field. Elementary examples are $\Bbb R$, $\Bbb C$, matrix algebras $M_n(k)$, polynomial algebras $k[X]$, etc., all of which are vector spaces by definition, and since there is multiplication in algebras, there are powers as well, assuming (power) associativity. – Ennar Jan 01 '17 at 01:20
  • I didn't find anything that shows what you say , I don't think you get the definitions – Seginus Jan 01 '17 at 01:26
  • @Seginus, I've just written you many examples, I'm not sure what you don't understand about them. – Ennar Jan 01 '17 at 01:27
  • when we're talking about power of an element this element isn't seen as a vector it is seen as an element over a ring the example you just gave right now are taken from an algebra and not in a vector space , so your definitions aren't correct – Seginus Jan 01 '17 at 01:30
  • @Seginus, algebra (over a field) = vector space + ring by definition, if I take power inside algebra, I'm taking a power of a vector, not much more to say about it. Contemplate on it. – Ennar Jan 01 '17 at 01:34
  • why you don't get it , taking a vector space which is not an algebra you can't talk about the power of the vectors , that's it , as easy as that – Seginus Jan 01 '17 at 01:37
  • @Seginus, I will repeat my comment: "Seginus, no, that is untrue, you can take powers of vectors in any algebra over a field." This effectively makes your statement "[... ] you're saying that you can speak about the power of a vector which is immpossible" invalid. Especially since in this case we are talking about real numbers, a well known algebra. – Ennar Jan 01 '17 at 01:40
  • OUUUUUUUUUUUPS i'm so sorry i'm stupid – Seginus Jan 01 '17 at 01:48
  • @Seginus, it's fine as long as we reached an agreement. Take care. – Ennar Jan 01 '17 at 01:55

2 Answers2

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Let $x_0\neq 0$ be a number so that $f(x_0)=0$. Then for any $c\in \Bbb R$, we have $f(cx_0)=cf(x_0)=0$.

Arthur
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we have $f(1^n)=nf(1)$ for all positive $n$, so $f(1)=0$. We also have $0=f((-1)^2)=2f(-1)$ and so $f(-1)=0$.

Since the function is linear and has two roots we have $f(x)=0$ for all $x$.

djechlin
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Asinomás
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