Some more proof problems (from CMI)....

1) If f(x)=xnn!+x(n1)(n1)!...x+1f(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.

Check out my set here

Note by Vishnu C
4 years, 1 month ago

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

This implies that f(a)f(a)=0    an1(n1)!=0    a=0 f(a) - f'(a) = 0 \implies \dfrac{a^{n-1}}{(n-1)!} = 0 \implies a = 0 . But 0 0 is obviously not a root of f(x) f(x) .Therefore, there exists no repeated root of f(x) f(x) .

For 2, you can see that modulo 2, f(a)f(1),f(0)1 f(a) \equiv f(1) , f(0) \equiv 1 ( depending on whether a is odd or even). for all integral a a . But, if a integral root b b did exist, that would imply that f(b)=00 f(b) = 0 \equiv 0 , which leads to a contradiction

Siddhartha Srivastava - 4 years, 1 month ago

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

vishnu c - 4 years, 1 month ago

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  1. We start by inducting on nn. Let fn(x)=k=0nxkk!f_n(x) = \sum_{k=0}^n \frac{x^k}{k!}
Let's see the base case, where n=1n=1 f1(x)=1+x=0x=1f_1(x) = 1+x =0 \Rightarrow x=-1 f1(x)f_1(x) has only 11 root and thus does not have repeated roots.

Let us assume that fn(x)=1+x+x22!++xnn!f_n(x) = 1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!} has no repeated roots.

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

fn+1(x)=1+x+x22!++xnn!+xn+1(n+1)!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 fn+1(x)f_{n+1}(x) fn+1(x)=1+x+x22!++xnn!f_{n+1}^{\prime}(x) = 1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}

Clearly, fn+1(x)f_{n+1}^{\prime}(x) has no repeated roots and must have nn distinct roots by our inductive hypothesis. Therefore, by Rolle's Theorem, we must have n+1n+1 distinct roots of fn+1(x)f_{n+1}(x). And thus proved.

Kishlaya Jaiswal - 4 years, 1 month ago

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@arian tashakkor

vishnu c - 4 years, 1 month ago

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

Arian Tashakkor - 4 years, 1 month ago

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I've posted some more proof notes.

vishnu c - 4 years, 1 month ago

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The first one can be done with the help of LMV or Rolles i guess??

Kunal Gupta - 4 years, 1 month ago

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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 xn/n!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.

vishnu c - 4 years, 1 month ago

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