and PlanetMath.org.I show that any polynomial of odd degree must have at least one root that is real.
Lemma 1:
lim (x → ∞) c/x = 0
where c is a nonzero constant
NOTE: This means that for any positive number ε, we can find a number δ such that:
for all abs(x) ≥ δ, abs(c/x) is less than ε
[See Definition 1, here for definition of mathematical limits]
Proof:
(1) Let ε be any positive number
(2) Assume c is positive.
(3) Let δ = 1 + c/ε
(4) So, then δ is greater than c/ε
(5) Which means that 1/δ is less than ε/c
(6) Which means that c/δ is less than ε
(7) For all n ≥ δ, it follows that:
c/n ≤ c/δ [which is less than ε from step #5 above]
(8) -c/δ is greater than -ε [this follows directly from step #5 from multiplying -1 to both sides]
(9) For all n ≤ -δ, it follows that:
c/n ≥ -c/δ [which is greater than -ε from step #7 above]
(10) Assume that c is negative
(11) Let δ = c/ε - 1
(12) So δ is less than c/ε
(13) So 1/δ is greater than ε/c
(14) Since c is negative, c/δ is less than ε [from multiplying a negative number to both sides]
(15) So for all n ≥ δ, it follows that:
c/n ≤ c/δ [which is less than ε from step #14 above]
(16) -c/δ is greater than -ε [this follows directly from step #14 from multiplying -1 to both sides]
(17) For all n ≤ -δ, it follows that:
c/n ≥ -c/δ [which is greater than -ε from step #16 above]
QED
Corollary 1.1:
limit (c/xi) = 0
where c is a constant and i is a positive integer
Proof:
(1) For i=1, it is true from Lemma 1 above
(2) Let δ be the same δ from Lemma 1 above which depends on the sign of c.
(3) For all n ≥ δ, we have:
c/ni ≤ c/n ≤ c/δ which is less than ε
(4) For all -n ≤ -δ, we have:
c/ni ≥ c/n ≥ -c/δ which is greater than -ε
QED
Corollary 1.2:
Let g(x) = a1/x + a2/x2 + ... + an-1/xn
limit (x → ∞) g(x) = 0
Proof:
(1) For each ai/xi, the limit is 0. [From Corollary 1.1 above]
(2) Since g(x) is the sum of these values, we apply the Addition Rule [see Corollary 8.1, here] to get:
limit (x → ∞) g(x) = 0
QED
Lemma 2:
If f(x) is a polynomial of odd degree, there exists values a,b such that:
f(a) is less than 0 and f(b) is greater than 0
Proof:
(1) Let f(x) be a polynomial of an odd degree n such that:
f(x) = a0xn + a1xn-1 + ... + an-1x + an = 0
(2) We can make f(x) monic (where xn does not have a coefficient) by dividing both sides by a0.
So, we can assume the following monic form:
xn + a1xn-1 + ... + an-1x + an = 0
(3) Let g(x) = (1/xn)[a1xn-1 + ... + an-1x + an] =
= a1/x + a2/x2 + ... + an-1/xn-1 + an/xn
(4) Then:
f(x) = xn[1 + g(x)]
(5) Using Corollary 1.2 above, we know that there exists δ such that:
for all abs(x) ≥ δ, abs(g(x)) is less than 1 [where ε = 1]
(6) So, it follows that 1 + g(x) is greater than 0.
(7) Let b be any positive number greater than δ.
(8) It follows that f(b) is greater than 0 since bn*[1 + g(x)] is greater than 0.
(9) Let a be any negative number where a is less than -δ.
(10) It follows that f(a) is less than 0 since an*[1 + g(x)] is less than 0.
QED
Theorem 3: A polynomial of odd degree has at least one real root
Proof:
(1) Let f(x) be a polynomial of odd degree.
(2) All polynomials are continuous [see Corollary 1.1, here], so f(x) is continuous.
(3) There exists a number a such that f(a) is less than 0. [see Lemma 2 above]
(4) There exists a number b such that f(b) is greater than 0. [see Lemma 2 above]
(5) Therefore, there exists at least one real root [see Weierstrass Intermediate Value Theorem, here]
QED
References
- "polynomial equation of odd degree", PlanetMath.Org
- Edwards, Jr. C. H. & Penney, David E., Calculus and Analytic Geometry, Prentice Hall, 1990
