Are there other methods for this question other than contradict the double root hypothesis?
Polynomial: prove there is no double root
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polynomials
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1The text in the img you link would have been shorter than your text, wouldn't it? – 2017-02-06
3 Answers
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Suppose $x_0$ is a double root for $E(x)$, this would imply $E(x_0)=E'(x_0)=0$. Note that $E'(x)=1+x+\frac{x^2}{2}+\frac{x^3}{3} = E(x)-\frac{x^4}{24}$. So if $x_0$ is a double root of $E(x)$ it follows $x_0=0$, but then $E(x_0) \neq 0$, absurd.
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Suppose that $E$ has the double root §x_0=$. Then
$E(x_0)=0$ and $E'(x_0)=0$.
Its your turn to obtain a contradiction from this.
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0any there any other methods than this? – 2017-02-11
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A multiple root of a polynomial $f(X)$ is the same as a common root fo $f(X)$ and $f'(X)$, or a root of $\gcd(f(X),f'(X))$. The $\gcd$ can be computed using Euclid's algorithm. In particular, if $f(X)$ is of the form $$f(x)=\sum_{k=0}^N\frac{x^k}{k!},$$ we have $f'(X)=f(X)-\frac{x^N}{N!}$, hence any common root is also a root of $\frac {x^N}{N!}$, i.e., $x=0$ is the only candidate - which is of course not a root of $f(X)$.