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Let $P(x)$ be a polynomial of degree $101$. Then $x\mapsto P(x)$ cannot be a one-one onto mapping, i.e., bijective function from $\Bbb{R}$ to $\Bbb{R}$. True or false?

I think is when we take $P(x)=P(y)$, we get : $$ ax^{101}+bx^{100}+\ldots+cx+d = ay^{101}+by^{100}+\ldots+cy+d $$ or $$ a\left(x^{101} - y^{101}\right)+b\left(x^{100} - y^{100}\right)+\ldots+c(x-y) = 0 $$ or $$ (x-y)\cdot(\textrm{some equation in }x,y) = 0. $$ Thus we get either $x=y$ or some other expression other than $x=y$.

As a result, $P(x)$ can't be one-one as had it been so, we should have got only $x=y$ as the solution to $P(x)=P(y)$.

Can you please let me know if I am doing it right. If not, what's the mistake I am making?

To check if $P(x)$ is onto, we need to write $y=P(x)$ and then find the expression in terms of $x$ and see if the range is equal to the co-domain. But I am not able to visualize how to check the surjectivity of the function. Can someone help me check if $P(x)$ is onto or not? I am a bit confused here. Thanks for all the help.

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A degree 101 polynomial could be a bijective function, but it doesn't have to be. It depends on which polynomial you choose. If the question asks "Can a degree 101 polynomial be a bijective function $\mathbb{R \to R}$?" then the answer is yes and you would prove that to be the correct answer by giving an example of a polynomial that has degree 101 and is a bijective function. –  Jim May 10 '13 at 22:45
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HITN: If you're doing this on the reals, then look at $P(x)=x^{101}$ –  fgp May 10 '13 at 22:45
    
ok..got it now..thanks to both of you for the feedback :) –  under-root May 10 '13 at 22:53
    
1 more clarification.What if the degree would have been 100 instead of 101??? –  under-root May 10 '13 at 22:57
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Polynomials of even degree can never be injective or surjective. As $\left|x\right|$ becomes large, $a_n x^n$ is the only term that really matters in $p(x) = a_n x^n + \ldots + a_0$, and when $n$ is even, $a_n x^n = a_n (-x)^n$, so $\lim_{x\to+\infty} p(x) = \lim_{x\to-\infty} p(x) = \operatorname{sign} (a_n)\infty$ (if you'll pardon my notation), and because polynomials are continuous, they are bounded on non-infinite intervals. Thus, as $x\to\pm\infty$, $p(x)\to\infty$ assuming the leading coefficient is positive), and $p(x)$ can never become unbounded negatively, because it is continuous. –  Stahl May 10 '13 at 23:00
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5 Answers

up vote 10 down vote accepted

$P(x) = x^{101}$ is a bijective polynomial of degree $101$. $Q(x) = x^{101}-x^2$ is of degree $101$, too, but it's not injective. So unless there's more to the question, an arbitrary 101st degree polynomial can be bijective, but not all of them are.

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Yes, the statement in the question is false. As was said in the comments, polynomials of even degree are never injective or surjective, and polynomials of odd degree are always surjective and may or may not be injective. –  Javier Badia May 10 '13 at 23:25
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yes...got it all now...thanks a lot for your suggestions. –  under-root May 10 '13 at 23:30
    
@sunnyverma: If the answer helped you, consider clicking the "accept" button next to it. –  Javier Badia May 11 '13 at 20:06
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A polynomial $P:\mathbb{R}\to\mathbb{R}$ is bijective if and only if $P'(x)$ never changes sign.

In less mathematical language, we need the polynomial to always go up, or always go down, and it's only allowed to level out momentarily.

With this in mind, it is fairly easy to write down a polynomial of odd degree that is bijective: only allow positive coefficients and terms of odd degree. For example, consider $P(x)=x^{101}+3x^{35}+4x$. You will find that its derivative is always positive.

It is not true that all bijective polynomials have the above form, as you can also check that $P(x)=x^{101}-x^2+10x$ also has an everywhere positive derivative.

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did not know this fact.thanks for the enlightment. –  under-root May 10 '13 at 23:22
    
+1 no qualms now! You could even strengthen it here, and say any polynomial is bijective iff $p'$ never changes sign, as that condition automatically rules out polynomials of even degree. –  Stahl May 10 '13 at 23:57
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Although (as Javier demonstrated) not all polynomials of odd degree are injective, they are all surjective. Say $p(x) = a_n x^n + \ldots + a_0$, where $n$ is odd. Then if $a_n > 0$, we will have $\lim_{x\to+\infty} p(x) = +\infty$ and $\lim_{x\to-\infty} p(x) = -\infty$. Since polynomials are continuous, we have to have hit everything in between somewhere along the way, so $p(x)$ is surjective (if $a_n < 0$, then $\lim_{x\to+\infty} p(x) = -\infty$ and $\lim_{x\to-\infty} p(x) = +\infty$). This happens because as $\left|x\right|$ becomes large, the highest degree term starts to contribute more and more compared to the other terms, so eventually it overwhelms them and the polynomial will simply take the sign of $a_n x^n$.

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Thus the easy half of the Fundamental Theorem of Algebra! –  Jp McCarthy May 10 '13 at 23:49
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A general version of this problem was addressed in this question on MO. We can state the question as follows:

Suppose $p$ is a polynomial of degree $101$ and real coefficients. What conditions must $p$ satisfy to ensure that $p$ is bijective?

As Stahl indicates in their answer, any polynomial of odd degree is surjective (using the intermediate value property). The question is then whether the polynomial is injective.

If $a<b$ and $p(a)=p(b)$, then by Rolle's theorem, there is a point $c$ in between, $a<c<b$, such that $p'(c)=0$. Hence, in order for $p$ to be injective, we need that $p'$ is never zero, that is, we need to understand when it is the case that a polynomial $q$ of degree $100$ and real coefficients (in this case, $q=p'$) has no real roots. This is precisely the question asked on MO. The selected answer actually tells you more: It gives you an algorithm for identifying precisely how many real roots a polynomial will have, without having to compute them.

Details are in the given link and references there. Briefly: You associate to $q$ a symmetric matrix $H$, called its Hermite form. The number of real roots of $q$ is the difference between the number of positive and negative eigenvalues of $H$ (the signature of $H$). To find this number, one computes the characteristic polynomial $c$ of $H$, because the eigenvalues of $H$ are the roots of $c$, and applies to it Descartes's rule of signs, which precisely tells us the number of positive and negative roots of $c$.

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thanks for your valuable feedback. –  under-root May 12 '13 at 11:18
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Just this past week, functionality came out in WolframAlpha allowing you to ask whether a function is bijective, injective, or surjective. We can apply this to your bijective example like so:
"is p(x) = x^101 bijective".

enter image description here

Or, "is x^101 - x^3 bijective", which has a prettier graph, too.

enter image description here

Not surprisingly, this works pretty well for polynomials or even for algebraic functions. You can certainly find transcendental functions that stump it, however.

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this is awesome stuff!!!thanks a lot for sharing it :))) –  under-root May 12 '13 at 11:16
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