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We know that $a_nx^n + a_{n-1}x^{n-1} + \dots + a_0 = b_nx^n + b_{n-1}x^{n-1} + \dots + b_0 $. How we can prove that $a_n = b_n , a_{n-1} = b_{n-1} , \dots ,a_0 = b_0$ . Also if in right side instead of $x$ we put $z$ is this statement true ? $a_n = b_n , a_{n-1} = b_{n-1} , \dots ,a_0 = b_0$

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    You need some quantifiers. If you mean $a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0 = b_n x^n + b_{n-1} x^{n-1} + \ldots + b_0$ **for every** $x$ (either in the real or complex numbers), then the answer is yes. In fact, you just need it to be true for at least $n+1$ different numbers. – 2017-02-23
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    Move everything to one side, and group the coefficients of all of the $x^k$ terms for each $k\in\{1,...,n\}$. Then use the fact that $x^k$ and $x^j$ are linearly independent for $j\neq k$. – 2017-02-23
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    @Dave Can you explain the last sentence ? – 2017-02-23
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    If the coefficients are from a finite field then the statement is false – 2017-02-23
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    For instance, $x^3$ and $x^2$ are linearly independent where $x$ is an indeterminate. So if you have an equation $c_1x^3+c_2x^2=0$ then we must have $c_1=c_2=0$. – 2017-02-23
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    @Dave So if $ax^2 + bx + c = 0$ then we must have $a=b=c=0$ ?! – 2017-02-23
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    If it is to hold for every $x$, then yes. – 2017-02-23

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That depends.

  • If you consider them as formal polynomials, then the equality of coefficients is by definition.

  • If you consider them as functions $\mathbb R\to\mathbb R$, you can insert $n+1$ different values for $x$ to obtain a set of $n+1$ independent linear equations for the $n+1$ variables $b_k$. Such a system has only a single solution, and it is obvious that $b_k=a_k$ is a solution.

  • If you consider them as functions acting on some other field, then the claim may be false. For example in $\mathbb Z/2\mathbb Z$, $x=x^2=x^3=x^4=\ldots$, therefore as functions in that field, all you can say is that $a_0 = b_0$ (obtained by inserting $0$) and $\sum_{k=0}^n a_k = \sum_{k=0}^n b_k$ (obtained by inserting $1$).

  • If you consider $x$ not as variable taking arbitrary values, but as a single specific value, then the claim is definitely false. Rather, you've got just a single equation that the coefficients have to fulfill.

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    For example when we can say $f(x) = \frac{ax+b}{cx+d}$ and its inverse are equal ? – 2017-02-23
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    @S.H.W: Over which field? – 2017-02-23
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    In the real numbers – 2017-02-23
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    @S.H.W: Then it's the second case in the list above. Thus $a=c$ and $b=d$. – 2017-02-23
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    But in the my book said $a = -d$ – 2017-02-23
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    Note : $f^{-1}(x) = \frac{-dx + b}{cx - a}$ – 2017-02-23
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    @S.H.W: Well, then "inverse" in that case doesn't mean "multiplicative inverse" (i.e. $1/f(x)$) but "function inverse" (that is, going from the function result back to the function argument). I misinterpreted because of the context; the *function inverse* does not mean that the polynomials in the numerator and denominator are equal. (Actually also for my interpretation I wasn't completely right, as not only $1$ but also $-1$ is its own multiplicative inverse, therefore there's a second solution). – 2017-02-23
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    So $a = -d$ is only solution ? – 2017-02-23
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    @S.H.W: It is the only *condition*. There are actually infinitely many solutions: You can choose $b$, $c$ and $d$ (almost) arbitrary, and then $a$ is fixed by the condition. OK, all solutions which only differ by a common factor for all four coefficients are equal, so effectively you've got only two free parameters, but that still gives infinitely many solutions. But anyway, all that is at best tangentially related to your question. – 2017-02-23
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    Okay , thank you a lot. – 2017-02-23
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    @S.H.W: I just noticed that I overlooked one other solution, that's quite obvious: $a=d,b=0,c=0$ (this gives $f(x)=x$, which clearly is invertible). But that plus all cases with $a=-d$ now definitely covers all solutions (you must have $a^2=d^2$, and if $a\ne-d$, then you must have $b=c=0$). – 2017-02-24
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    Okay , but $f(x) = \frac{ax+b}{cx+d}$ is homographic function and the condition is $c\not= 0$. – 2017-02-24
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    @S.H.W: Well, that was a condition you didn't mention. – 2017-02-24
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    Yes you are right . So the only solution is $a = -d$ . Is this right ? – 2017-02-24
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    Yes, as I wrote, if $a\ne -d$, then it follows that $c=0$, which you said was excluded. Therefore $a=-d$ remains as only option. – 2017-02-24
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    Okay , thank you a lot for your answers . – 2017-02-24