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

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?

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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

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 $f(c)=0$.

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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 $4.4$.

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Hint:

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

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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 $(3)$.

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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 $g(x_0)=0$

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 $g(x_0)=0$.
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
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