Waste less time on Facebook — follow Brilliant.
×

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

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.

Check out my set here

Note by Vishnu C
1 year, 8 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

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, 8 months ago

Log in to reply

@Siddhartha Srivastava Ya. I used the exact same methods for both the questions. Great minds think alike ; ) Vishnu C · 1 year, 8 months ago

Log in to reply

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

Let 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, 8 months ago

Log in to reply

The first one can be done with the help of LMV or Rolles i guess?? Upanshu Gupta · 1 year, 8 months ago

Log in to reply

@Upanshu Gupta 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. Vishnu C · 1 year, 8 months ago

Log in to reply

@arian tashakkor Vishnu C · 1 year, 8 months ago

Log in to reply

@Vishnu C ok these ones are somewhat hard to understand I should say ... Arian Tashakkor · 1 year, 8 months ago

Log in to reply

@Arian Tashakkor I've posted some more proof notes. Vishnu C · 1 year, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...