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.

Prove that the equation $x^{10000} + x^{100} - 1 = 0$ has a solution with $0 < x < 1$.

This is a homework question. I know I could probably find a solution that would complete the proof, but I don't think that is what this question is asking. What proof-techniques should I use to prove this is true?

share|improve this question
7  
Why do you say you could probably find a solution? Do you have any idea about how you might do that? –  TonyK Dec 5 '11 at 14:54

5 Answers 5

up vote 38 down vote accepted

I presume your teacher wants you to use the Intermediate Value Theorem:

Let $f(x)=x^{10000}+x^{100}-1$. Then $f$ is continuous on $[0,1]$, $f(0)=-1 $ and $f(1)=1 $.

Since $f(0)=-1<0<1=f(1)$, the Intermediate Value Theorem guarantees that there is a point $c$ in the interval $[0,1]$ with $f(c)=0$. Since that point can't be $0$ or $1$, it must be in $(0,1)$.



Informally: Since $f$ is continuous over $[0,1]$, its graph is "unbroken" over $[0,1]$. Since $f(0)=-1$ and $f(1)=1$, the graph of $f$ must cross the horizontal line $y=0$ somewhere over the interval $(0,1)$. This intersection point gives you a value $c$ in $(0,1)$ where $f(c)=0$.

share|improve this answer
4  
IVT is easily one of the most under-appreciated theorems out there. –  joshin4colours Dec 5 '11 at 18:59

Just for fun, fairly elementary methods can give much better bounds. Let $x = 1 - \frac{y}{10000}$: then we want to solve $$\left( 1 - \frac{y}{10000} \right)^{10000} + \left( 1 - \frac{y}{10000} \right)^{100} - 1 = 0$$

where we know that

$y \in (0, 10000)$.

Actually $y$ lies in a much smaller interval; indeed using the inequality $(1 - t)^n \le e^{-nt}$ (which follows by convexity) we see that $$e^{-y} + e^{- \frac{y}{100} } - 1 \ge 0$$

and in particular $2 e^{ - \frac{y}{100} } - 1 \ge 0$, so actually

$y \in (0, 100 \ln 2) \subset (0, 70)$.

Now, using the inequality $e^{-x} \le 1 - \frac{x}{2}$ for $x \in [0, 1]$ (which also follows by convexity) we have $$e^{-y} + 1 - \frac{y}{200} - 1 \ge 0$$

which gives $$e^{-y} \ge \frac{y}{200}.$$

Since $e > 2$ we have $e^{-6} < \frac{1}{64} < \frac{6}{200}$, so

$y \in (0, 6)$.

Since $e^2 > 6$ we have $e^{-5} < \frac{1}{72} < \frac{5}{200}$, so

$y \in (0, 5)$.

In fact the actual value of $y$ is approximately $4.4$.

share|improve this answer

Hint:

If $f(a)*f(b)<0$ this implies there's at least one root in the interval $[a,b]$.

Reference.

share|improve this answer

I'm trying to improve Qiaochu Yuan's bounds.

The equation $$x^{100}+x^{10\thinspace000}=1 \qquad\qquad(1)$$ obviously has exactly one solution $\xi\in\ ]0,1[\ $. We put $100=:n$ and $$x:=1-{y\over n^2}$$ with a new unknown $y>0$. In this way equation $(1)$ becomes $$\Bigl(f(y):=\Bigr)\quad1-\Bigl(1-{y\over n^2}\Bigr)^n=\Bigl(1-{y\over n^2}\Bigr)^{n^2}\quad\Bigl(=: g(y)\Bigr)\ .\qquad\qquad(2)$$ In order to get some indication on the order of magnitude to be expected we assume $y\ll n$ and then have approximatively $$f(y)\doteq {y\over n} \>, \quad g(y)\doteq e^{-y}\ .$$ In this way equation $(2)$ morphs into the simple equation $y\, e^y=100$ with the solution $y_*\doteq 3.3856$. It will turn out that this value is a very good approximation to the true solution $\eta$ of $(2)$.

The function $f$ is monotonically increasing for $0<y<n^2$, and the function $g$ is monotonically decreasing there. In the following we shall consider the two $y$-values $y_1:=3.38$ and $y_2:=3.40$, and (with the help of a pocket calculator) we shall prove that $$f(y_1)<g(y_1)\>, \quad f(y_2)> g(y_2)\ .\qquad\qquad(3)$$ This implies $y_1<\eta<y_2$, resp. $$1-0.000340<\xi<1-0.000338\ .$$

So we need bounds for $f$ and $g$. Concerning $f$ it is easily seen that $${y\over n}\Bigl(1-{y\over 2n}\Bigr)<f(y)<{y\over n}\qquad (0<y< n^2)$$ – the upper bound being nothing else but Bernoulli's inequality. This immediately implies $$f(3.38)<0.0338\>,\qquad f(3.40)>0.0340\,(1-0.0170)=0.03342\ .$$ For $g$ we begin with $$\log(1-t)=-\Bigl(t+{t^2\over2}+{t^3\over3}+\ldots\Bigr)\ \cases{\ <-t \cr \ >-t -{\displaystyle{t^2/2\over 1-t}} \cr}\qquad (0<t<1)\ .$$ Putting $t:={\displaystyle{y\over n^2}}$ and multiplying with $n^2$ we get the estimates $$-y -{y^2 \over 2(n^2-y)}< \log \bigl(g(y)\bigr)< -y\ ,$$ and using the inequality ${\displaystyle{\exp\Bigl({-y^2\over 2(n^2-y)}\Bigr)>1-{y^2\over 2(n^2-y)}}}$ we therefore obtain $$e^{-y}\Bigl(1-{y^2\over 2(n^2-y)}\Bigr)< g(y)< e^{-y}\ .$$ This implies $$g(3.38)> 0.034047\,(1-0.000571)=0.034028\>,\quad g(3.40)<0.03337\ .$$ It follows that the values of $f$ and $g$ at the places $y_1$ and $y_2$ obey the stated relation $(3)$.

share|improve this answer

Let $g(x)=x^{10000}+x^{100}-1$. Obviously,$g(x)$ is continuous and increasing in $[0,+\infty)$, because $g(0)=-1 \ and \ g(1)=1$, so $x_0\in (0,1)$ if $g(x_0)=0$

Added:

For a function in the form of $f(x)=x^n, n\in \mathbb N$,we can see that $f(x)$ is increasing on$[0,+\infty)$.

For any $x>y, x,y\in (0,+\infty)$, we have $f(y)>0$, and $\frac{f(x)}{f(y)}=(\frac{x}{y})^n>1$, since $\frac{x}{y}>1$, therefore, we get$f(x)>f(y)$. Also, for any$x\in(0,+\infty)$,$f(x)>f(0)=0$, so we can say $f(x)$ is increasing on $[0,+\infty)$.

Of course, the continuity tells you that there is at least one point $x_0$ satisfying $g(x_0)=0$, however, my conclusion is stronger: there is only one point $x_0$ satisfying $g(x_0)=0$.

share|improve this answer
    
of course,two points can't tell me that. However, I think a function in the form $f(x)=x^n,n\in \mathbb N$ is an –  Huang Dec 5 '11 at 23:14
    
I think I agree with what you were about to write before you got cut off. I just thought the phrasing was strange. –  Dylan Moreland Dec 5 '11 at 23:15
    
of course,two points can't tell me that. However, I think a function in the form $f(x)=x^n,n\in \mathbb N$ is elementary and it's monotonicity in $[0,\infty)$ is obvious. I am not familiar with Math jargons in English, but I can show it. –  Huang Dec 5 '11 at 23:20
    
@DylanMoreland: Sorry, I just typed my comment on my phone. You know, it's slower to type than on a computer. And I finished my comment without care. That's why I showed such a strange comment... –  Huang Dec 5 '11 at 23:21
    
I often type on a touch screen, so I understand! I hope it didn't seem like I was picking on you. And you're right: it's a sum of increasing functions. –  Dylan Moreland Dec 5 '11 at 23:46

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.