polynomial. (1 Viewer)

[seanieg89]

Prove that for all positive integers n, the polynomial:



has no real roots.

 

[barbernator]

First, Test for x=0. The function does not have a root at x=0

<a href="http://www.codecogs.com/eqnedit.php?latex=P(x)=\sum_{k=0}^{2n}\frac{x^{k}}{k!}\\\\ P'(x)=\sum_{k=0}^{2n-1}\frac{x^{k}}{k!}\\\\ P''(x)=\sum_{k=0}^{2n-2}\frac{x^{k}}{k!}\\\\ the~function~has~a~minimum~stationary~point~at ~\alpha \\\\ hence,~P'(\alpha )=0.\\\\ at~this~point,~P(\alpha )-P'(\alpha )=P(\alpha )=\frac{\alpha ^{2n}}{(2n)!}~which~is~positive.\\\\ hence,~as~the~function~is~of~an~even~degree,~P(x)~tends~towards~@plus;\infty ~for~x->\pm \infty \\\\ hence,~as~all~minimum~stationary~points~are~>0~the~function~has~no~real~roots." target="_blank"><img src="http://latex.codecogs.com/gif.latex?P(x)=\sum_{k=0}^{2n}\frac{x^{k}}{k!}\\\\ P'(x)=\sum_{k=0}^{2n-1}\frac{x^{k}}{k!}\\\\ P''(x)=\sum_{k=0}^{2n-2}\frac{x^{k}}{k!}\\\\ the~function~has~a~minimum~stationary~point~at ~\alpha \\\\ hence,~P'(\alpha )=0.\\\\ at~this~point,~P(\alpha )-P'(\alpha )=P(\alpha )=\frac{\alpha ^{2n}}{(2n)!}~which~is~positive.\\\\ hence,~as~the~function~is~of~an~even~degree,~P(x)~tends~towards~+\infty ~for~x->\pm \infty \\\\ hence,~as~all~minimum~stationary~points~are~>0~the~function~has~no~real~roots." title="P(x)=\sum_{k=0}^{2n}\frac{x^{k}}{k!}\\\\ P'(x)=\sum_{k=0}^{2n-1}\frac{x^{k}}{k!}\\\\ P''(x)=\sum_{k=0}^{2n-2}\frac{x^{k}}{k!}\\\\ the~function~has~a~minimum~stationary~point~at ~\alpha \\\\ hence,~P'(\alpha )=0.\\\\ at~this~point,~P(\alpha )-P'(\alpha )=P(\alpha )=\frac{\alpha ^{2n}}{(2n)!}~which~is~positive.\\\\ hence,~as~the~function~is~of~an~even~degree,~P(x)~tends~towards~+\infty ~for~x->\pm \infty \\\\ hence,~as~all~minimum~stationary~points~are~>0~the~function~has~no~real~roots." /></a>

im not sure whether my justification is perfect...

 

[Fus Ro Dah]

Personally, I would use an inductive proof.

 

[Kiraken]

Personally, I would use an inductive proof.

Yep. I think i've seen this question in one of the textbooks before....

 

[seanieg89]

Your reasoning is essentially correct barb, although there are some unstated results that you used implicitly. (Why must there exist a stationary point which is a local minimum? And why must the poly attain a global minimum at such a local min?) I personally used an inductive approach.