Simple zeros ain't so simple

Calculus Level 4

For positive integer $n$ , how many roots of $x$ in the interval $-1\leq x\leq1$ does $\dfrac{d^n}{dx^n} (x^2-1)^n =0$ have?

n

{n}^{2}

2n

Impossible to know

n-1

Infinitely many

2 solutions

Rishabh Deep Singh
Jan 24, 2016

let n=3 then differentiate it 3 times equation becomes $x\left( 5{ x }^{ 2 }-3 \right) =0\\ x=0\quad ;\quad x=\pm \sqrt { \frac { 3 }{ 5 } } \approx \pm 0.77$

hence 3 roots.

Moderator note:

This solution doesn't prove the result.

You can't choose a value for $n$ . You have to prove for all $n$ .

Lucas Tell Marchi - 5 years, 4 months ago

to get the answer we can put n= 3

Rishabh Deep Singh - 5 years, 4 months ago

that doesn't give you that it's the case for all numbers,although the option "it depends" wasn't on there you still should have a proof that this applies to all natural numbers for a valid solution,but it`s enough to get this question right

Hamza A - 5 years, 4 months ago

@Hamza A – This is a higher level problem hat i didn't studied yet so i tried to solve it by substituting n=3

Rishabh Deep Singh - 5 years, 4 months ago

@Rishabh Deep Singh – you have a completely valid answer and your reasoning is sound for this problem,and it's ok if you don't know everything about the problem ,that's why we're all here

to learn (:

@Lucas Tell Marchi 's solution was nice ,i learned something today!

Hamza A - 5 years, 4 months ago

Lucas Tell Marchi
Jan 25, 2016

Well, using Legendre polynomials:

$\frac{\mathrm{d}^{n} }{\mathrm{d} x^{n}} (x^{2} - 1)^{n} = 2^{n} n! P_{n}(x)$

Where $P_{n}(x)$ is the $n$ -th Legendre polynomial with domain ranging from $-1$ to $1$ . Therefore

$\frac{\mathrm{d}^{n} }{\mathrm{d} x^{n}} (x^{2} - 1)^{n} = 0 \Rightarrow P_{n}(x) = 0$

And as $P_{n}(x)$ is a polynomial of degree $n$ , it has $n$ complex roots.

Your solution doesn't yet prove that we have $n$ roots in the interval $[-1, 1 ]$ . All that we know is it has $n$ complex roots, why must all of them be in that interval?

How are you using the fact that we have a Legendre polynomial?

Calvin Lin Staff - 5 years, 4 months ago

The Legendre polynomials are only defined on this interval because the argument of the polynomial is $\cos(\theta)$ which we call $x$ for short. As it is a polynomial, it must have $n$ complex roots in its domain. And what I used is a property of Legendre polynomials. If $x$ ranges on this domain, my first equation is always true.

Lucas Tell Marchi - 5 years, 4 months ago

Not true. The Legendre polynomials are defined over all of $\mathbb{C}$ . For example, $P_1 (x) = x, P_2 (x) = \frac{1}{2} ( 3x^2 - 1 )$ . Yes, they arise from thinking about $\cos \theta$ , but just because you put in a restriction doesn't imply that the polynomial must behave nicely as you want it to.

However, it is true that "the Legendre polynomial has $n$ roots in the interval $[-1,1]$ , which you have not explained / stated in your solution. That is the key to concluding that "there are $n$ roots in the interval $[-1,1]$ " as opposed to "there are $n$ complex roots" which is the final statement in your solution.

Calvin Lin Staff - 5 years, 4 months ago

you said n complex roots but at n = 3 in my case there are 3 real roots

Rishabh Deep Singh - 5 years, 4 months ago

Real numbers are complex numbers with no imaginary part.

Lucas Tell Marchi - 5 years, 4 months ago

Simple zeros ain't so simple

2 solutions

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