Polynomial bounded w.r.t supremum norm

[Somefantastik]

Homework Statement

E₁ = {p_n(t) = nt(1-t)ⁿ:n in N};

E₂ = {p_n(t) = t + (1/2)t² +...+(1/n)tⁿ: n in N};

where N is set of natural numbers

is the polynomial bounded w.r.t the supremum norm on P[0,1]?

Homework Equations

supremum norm = ||*|| = sup{|p_n(t)|: t in [0,1]}

The Attempt at a Solution

I know that the set is bounded on P[0,1] if I can show that the supremum norm is less than some constant for all n. Can someone give me some advice on how to show either polynomial is bounded or not?

I think E₁ is unbounded, no matter your choice of t, since the polynomial's value will forever increase as n in increases. I feel the same can be said of E₂ but I know this is wrong.

 

[ystael]

You can answer this fairly directly, by computing the maximum and minimum values of each $$p_n$$; because polynomials are smooth functions, the extreme values of $$p_n$$ on $$[0,1]$$ must occur at a critical point of $$p_n$$ or at one of the endpoints.

The members of $$E_2$$ share a particular property which will save you even this computation.

 

[Somefantastik]

I'm not seeing the special property of E₂. Its min value must be at t = 0 for all n, and max when t = 1, but I can't pin down a value for sup. It looks like an expanded function but it doesn't look familiar.

factoring out a t makes p₂(t) = t(1 + t + (1/2)t + (1/3)t² + ... + (1/n)t^n-1), but I'm still not seeing the trick.

 

[Somefantastik]

Also, at E₁, the critical points are t = 0 and t = 1. At both of these points, no matter the value of N, the polynomial is zero. Does this mean that is my sup? That doesn't make sense to me b/c the shape of this function spikes in the middle between t = 0 and t = 1.

 

[Somefantastik]

for E₂, p'(t) gives the geometric series, which equals 1/(1-t) on the interval |t|<1. The second derivative is 1/(1-t)² < 0

What can I do with this information? The series doesn't converge at either t = 0 or t = 1.

 

[ystael]

Also, at E₁, the critical points are t = 0 and t = 1. At both of these points, no matter the value of N, the polynomial is zero. Does this mean that is my sup? That doesn't make sense to me b/c the shape of this function spikes in the middle between t = 0 and t = 1.

If the function is zero at 0 and 1, and positive in the middle, then there must be another critical point between, no?

For $$E_2$$, put $$q_n(t) = t^n/n$$, so that $$p_n(t) = \sum_{1\leq j\leq n} q_n(t)$$. Where does each $$q_n$$ reach its maximum absolute value? What does this tell you about the $$p_n$$?

 

[Somefantastik]

If the function is zero at 0 and 1, and positive in the middle, then there must be another critical point between, no?

yes, and it'll be a function of n, but as n -> inf, isn't the max value of the function going to diverge to inf as well?

For $$E_2$$, put $$q_n(t) = t^n/n$$, so that $$p_n(t) = \sum_{1\leq j\leq n} q_n(t)$$. Where does each $$q_n$$ reach its maximum absolute value? What does this tell you about the $$p_n$$?

q_n reaches its max value at t = 1 => $$p_n(1) = \sum_{1\leq j\leq n}q_n(1) = \sum_{1\leq j\leq n}\frac{1}{n}$$

so the value of this polynomial is the sum of this series and is finite and therefore this polynomial is bounded.

 

[ystael]

yes, and it'll be a function of n, but as n -> inf, isn't the max value of the function going to diverge to inf as well?

That's precisely what the question asks you to determine.

q_n reaches its max value at t = 1 => $$p_n(1) = \sum_{1\leq j\leq n}q_n(1) = \sum_{1\leq j\leq n}\frac{1}{n}$$
so the value of this polynomial is the sum of this series and is finite and therefore this polynomial is bounded.

Bounded? Uniformly in $$n$$? Look again.

 

[Somefantastik]

That's precisely what the question asks you to determine.

I understand that, but I am not confident about my answer.

Bounded? Uniformly in $$n$$? Look again.

you're right, I got mixed up with our indexes on our summation notation. Anyway, the sum of (1/n) to infinity is harmonic sequence which does not converge, and therefore THIS polynomial is not bounded. Am I getting close?