Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having a problem with this question. Can someone help me please.

We are defining a sequence of polynomials such that:

$P_0(x)=1; P_n'(x)=nP_{n-1}(x) \mbox{ and} \int_{0}^1P_n(x)dx=0$

I need to prove, by induction, that $P_n(x)$ is a polynomial in $x$ of degree $n$, the term of highest degree being $x^n$.

Thank you in advance

share|improve this question
    
Where are you stuck? –  Pragabhava Nov 8 '12 at 2:11
    
For n=0. It is obvious. I am having trouble to show the property for n+1 –  Carpediem Nov 8 '12 at 2:15
    
Suppose $P_{n+1}$ has leading term $ax^r$ for some $a$ and $r$. The defining equation and the induction hypothesis tell you what about $a$ and $r$? –  Gerry Myerson Nov 8 '12 at 2:17
add comment

1 Answer

up vote 1 down vote accepted

Recall that $\displaystyle \int x^n dx = \dfrac{x^{n+1}}{n+1}$. Hence, if $P_n(x)$ is a polynomial of degree $n$, then it is of the form $$P_n(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$$ Since $P_{n+1}'(x) = (n+1) P_n(x)$, we have that $$P_{n+1}(x) = \int_{0}^x (n+1) P_n(y) dy + c$$ Hence, $$P_{n+1}(x) = \int_{0}^x (n+1) \left(a_n y^n + a_{n-1} y^{n-1} + \cdots + a_1 y + a_0\right) dy + c\\ = a_n x^{n+1} + a_{n-1} \left(\dfrac{n+1}n\right) x^n + \cdots + a_{1} \left(\dfrac{n+1}2\right) x^2 + a_{0} \left(\dfrac{n+1}1\right) x + c$$ Now finish it off with induction.

share|improve this answer
    
We say to such sequences as an appeal polynomial sequences. –  hassan joulani Jul 2 '13 at 16:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.