1) If \(f(x)=\frac{x^n}{n!}+\frac{x^(n-1)}{(n-1)!}...x+1\) then prove that f cannot have repeated roots.

This one I did not expect it:

2) If f(x) is a polynomial with integer coefficients such that f(0)=1 and f(1)=5, then prove that f cannot have integer roots.

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TopNewestLet us assume that \(f_n(x) = 1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}\) has no repeated roots.

We now study the function \(f_{n+1}(x)\)

\[f_{n+1}(x) = 1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}+\frac{x^{n+1}}{(n+1)!}\]

Observe the derivative of \(f_{n+1}(x)\) \[f_{n+1}^{\prime}(x) = 1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}\]

Clearly, \(f_{n+1}^{\prime}(x)\) has no repeated roots and must have \(n\) distinct roots by our inductive hypothesis. Therefore, by Rolle's Theorem, we must have \(n+1\) distinct roots of \(f_{n+1}(x)\). And thus proved. – Kishlaya Jaiswal · 1 year, 5 months ago

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For 1., you can use the fact that if \( f(x) \) has a repeated root, then \( f'(x) \) has the same root, say \( a \) .

This implies that \( f(a) - f'(a) = 0 \implies \dfrac{a^{n-1}}{(n-1)!} = 0 \implies a = 0 \). But \( 0 \) is obviously not a root of \( f(x) \).Therefore, there exists no repeated root of \( f(x) \).

For 2, you can see that modulo 2, \( f(a) \equiv f(1) , f(0) \equiv 1 \)( depending on whether a is odd or even). for all integral \( a \). But, if a integral root \( b \) did exist, that would imply that \( f(b) = 0 \equiv 0 \), which leads to a contradiction – Siddhartha Srivastava · 1 year, 5 months ago

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– Vishnu C · 1 year, 5 months ago

Ya. I used the exact same methods for both the questions. Great minds think alike ; )Log in to reply

The first one can be done with the help of LMV or Rolles i guess?? – Upanshu Gupta · 1 year, 5 months ago

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– Vishnu C · 1 year, 5 months ago

The first one is pretty simple. I don't like using LMV or Rolle's theorem, or at least their terms in anything that can be solved with a rough sketch. Although I don't use it, my method is pretty much related to it. If the function has repeated roots, then it's derivative also has a root at that point. If that's the case, then the \(x^n/n!\) factor in the function has no consequence in altering the value taken by the function. In that case, x must have a repeated root at 0. But substituting 0 in the place of x gives you 1. So 0 is not a root and the function doesn't have any repeated roots.Log in to reply

@arian tashakkor – Vishnu C · 1 year, 5 months ago

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– Arian Tashakkor · 1 year, 5 months ago

ok these ones are somewhat hard to understand I should say ...Log in to reply

– Vishnu C · 1 year, 5 months ago

I've posted some more proof notes.Log in to reply