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Polynomial including Sigma Notation

PurpleDude

New member
Feb 7, 2014
10
Hello everyone!

I have this polynomial: $p(x) =$ \(\displaystyle 1 + \sum_{k=1}^{13}\frac{(-1)^k}{k^2}x^k\)

- I'm supposed to show that this polynomial must have at least one positive real root.

- I'm supposed to show that this polynomial has no negative real roots.

- And I'm supposed to show that if $z$ is any root of this polynomial, then $|z| < 170$

I do not know how to start this question, so any guidance on these three steps would be appreciated. :)
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Hello everyone!

I have this polynomial: $p(x) =$ \(\displaystyle 1 + \sum_{k=1}^{13}\frac{(-1)^k}{k^2}x^k\)

- I'm supposed to show that this polynomial must have at least one positive real root.
Substitute nonnegative values of x into p and see if the sign of p changes.

- I'm supposed to show that this polynomial has no negative real roots.
Check to see if there are any turning points to the left of the root you found in the first part.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
An easier way to show $p$ has a positive root:

Note that $p(0) = 1 > 0$ and that the leading term is $\dfrac{-1}{169} < 0$.

As $x \to +\infty$ we have $p(x) \to -\infty$, so $p$ must cross the $x$-axis somewhere to the right of 0, by continuity.

The bound of 170 is very crude, here is a hint on how to proceed:

Show that if $|x| \geq 170$ that:

$\left|\dfrac{x^{13}}{169}\right| > |x^{12}|$

and that:

$\left|\dfrac{x^k}{k^2}\right| < \left|\dfrac{x^{12}}{13}\right|$

for each $k = 1,2,\dots,12$ (I've left one term out....why?).

Use this to prove that for such $x$, we have $|p(x)| > 0$.
 

PurpleDude

New member
Feb 7, 2014
10
An easier way to show $p$ has a positive root:

Note that $p(0) = 1 > 0$ and that the leading term is $\dfrac{-1}{169} < 0$.

As $x \to +\infty$ we have $p(x) \to -\infty$, so $p$ must cross the $x$-axis somewhere to the right of 0, by continuity.

The bound of 170 is very crude, here is a hint on how to proceed:

Show that if $|x| \geq 170$ that:

$\left|\dfrac{x^{13}}{169}\right| > |x^{12}|$

and that:

$\left|\dfrac{x^k}{k^2}\right| < \left|\dfrac{x^{12}}{13}\right|$

for each $k = 1,2,\dots,12$ (I've left one term out....why?).

Use this to prove that for such $x$, we have $|p(x)| > 0$.
The parts where I was to prove the postive or negative roots made a lot of sense thank you, but can you re-explain showing that if z is any root of p(x), then |z| < 170
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
I haven't "explained", I've given a HINT. The idea is you're supposed to do something WITH that hint.

The basic idea is to show that for $|x| \geq 170$, that the leading term "dominates", that it's so large that it is larger than the sum of absolute values of all the other terms, and thus larger than the absolute value of the rest of the polynomial.

So if:

$\left|\dfrac{x^{13}}{169}\right| > |x^{12}| = 13\left|\dfrac{x^{12}}{13}\right|$

$\displaystyle > \sum_{k = 1}^{12} \left|\frac{(-1)^k}{k^2}x^k\right| + |1|$

$\displaystyle \geq \left|\sum_{k=1}^{12} \frac{(-1)^k}{k^2} + 1\right|$

then $|p(x)| > 0$

(this is because if:

$|a| > |b|$ then $|a + b| > 0$).

So left of -170 we are always above the x-axis, and right of 170 we are always below the x-axis.