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Show that α^6 = 3.

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anemone

MHB POTW Director
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Feb 14, 2012
3,683
Hi all,

This question seems easy but I have tried many different ways to find the proof, still I couldn't get through. Could someone please walk me through this problem?

Thanks in advance.

Question:

If \(\displaystyle \alpha\) is a real root of \(\displaystyle x^5-x^3+x-2=0\), show that \(\displaystyle \alpha^6=3\)
 

Opalg

MHB Oldtimer
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Feb 7, 2012
2,706
Re: Show that α^6=3.

If \(\displaystyle \alpha\) is a real root of \(\displaystyle x^5-x^3+x-2=0\), show that \(\displaystyle \alpha^6=3\)
I am suspicious about this result. If $x^6=3$ then $x^3=\sqrt3$, and the equation $x^5-x^3+x-2=0$ can be written $\sqrt3x^2 + x -(2+\sqrt3) = 0.$ The positive root of that quadratic is $\dfrac{-1+\sqrt{13+8\sqrt3}}{2\sqrt3}\approx 1.207331$ according to my calculator. But $3^{1/6}\approx 1.200937.$ Those numbers are close, but definitely not the same.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Re: Show that α^6=3.

Certainly $\alpha^6\neq 3$. In fact, if $\alpha=\sqrt[6]{3}$ is a real root, must verify: $$\alpha^5-\alpha^3+\alpha-2=\sqrt[6]{243}+\sqrt[6]{27}+\sqrt[6]{3}-2=0\quad (*)$$ This means that $\sqrt[6]{243},\sqrt[6]{27},\sqrt[6]{3},1$ are linearly dependent over $\mathbb{Q}$. But the minimal polinomial of $\sqrt[6]{3}$ is $x^6-3$ which implies that a basis of $[\mathbb{Q}(\sqrt[6]{3}):\mathbb{Q}]$ is: $$\{1,\sqrt[6]{3},\sqrt[6]{9},\sqrt[6]{27},\sqrt[6]{81},\sqrt[6]{243}\}$$ This is a contradiction with $(*)$.
 
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anemone

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Feb 14, 2012
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Re: Show that α^6=3.

Hmm...yes, you're right, Opalg, this problem doesn't sound right to me now and I must also admit that I have not been able to locate the original problem yet. :mad:

(Edit: Thanks to you too, Fernando as I have come to realize it by now that this problem has a flaw in it...)
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,683
Re: Show that α^6=3.

I'm very glad to be able to finally locate the original problem and frankly, I still need help on this problem...:eek:

Problem:
If \(\displaystyle \alpha\) is a real root of \(\displaystyle f(x)=x^5-x^3+x-2\), then prove that \(\displaystyle [\alpha^6]=3\) where \(\displaystyle [a]\) denotes the greatest integer function.
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Show that α^6=3.

I'm very glad to be able to finally locate the original problem and frankly, I still need help on this problem...:eek:

Problem:
If \(\displaystyle \alpha\) is a real root of \(\displaystyle f(x)=x^5-x^3+x-2\), then prove that \(\displaystyle [\alpha^6]=3\) where \(\displaystyle [a]\) denotes the greatest integer function.
Just as a note: Some may be more familiar with the equivalent "floor function" notation:

\(\displaystyle \left\lfloor \alpha^6 \right\rfloor=3\)
 

Petrus

Well-known member
Feb 21, 2013
739
Re: Show that α^6=3.

I'm very glad to be able to finally locate the original problem and frankly, I still need help on this problem...:eek:

Problem:
If \(\displaystyle \alpha\) is a real root of \(\displaystyle f(x)=x^5-x^3+x-2\), then prove that \(\displaystyle [\alpha^6]=3\) where \(\displaystyle [a]\) denotes the greatest integer function.
Hello anemone,
We only get one real root and rest complex :) citat from Opalg "the equation $x^5-x^3+x-2=0$ can be written $\sqrt3x^2 + x -(2+\sqrt3) = 0.$ The positive root of that quadratic is $\dfrac{-1+\sqrt{13+8\sqrt3}}{2\sqrt3}\approx 1.207331$"
and \(\displaystyle 1.207331^6 \approx 3.0971205\)
and the greatest integer function become \(\displaystyle [3.0971205] = 3\) (cause you round it down)

edit: I think this is wrong, I have no clue how you calculate the roots of the polynom, Ignore this and wait for someone else... (you got \(\displaystyle [\alpha^6]=3\) and that is prob not the same as \(\displaystyle \alpha^6=3\)
Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited:
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anemone

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Feb 14, 2012
3,683
Re: Show that α^6=3.

I have found the solution suggested by someone who had solved this problem correctly and I want to share it with the members of MHB here.

We're given the equation of the function of f of x, i.e. \(\displaystyle f(x)=x^5-x^3+x-2\) and that \(\displaystyle \alpha\) is the real root of \(\displaystyle f(x)\).

First, we differentiate the function w.r.t.x and get:

\(\displaystyle f'(x)=5x^4-3x^2+1=5\left(\left(x^2-\frac{3}{10}\right)^2+\frac{11}{100}\right)\)

i.e. \(\displaystyle f'(x)\ge0\) for all real \(\displaystyle x\). This means \(\displaystyle f(x)\) is monotonically increasing, thus, \(\displaystyle f(x)\) has one and only one root, i.e. \(\displaystyle \alpha\).

From \(\displaystyle f(0)=-2\) and \(\displaystyle f'(2^{\frac{1}{3}})=2^{\frac{5}{3}}+2^{\frac{1}{3}}-4>0\), we can conclude by the Intermediate Value Theorem that the only root lies in \(\displaystyle (0, 2^{\frac{1}{3}})\).

Therefore, we get \(\displaystyle 0<\alpha<2^{\frac{1}{3}}\).

Raise both sides of the inequality to the sixth power we get:

\(\displaystyle 0<\alpha^6<4\)

Next, if we can prove that \(\displaystyle 3<\alpha^6<4\), we can also conclude that \(\displaystyle [\alpha]3\).

There maybe plenty of ways to prove that \(\displaystyle \alpha^6>3\) but this is what I did for I couldn't be able to fathom the working of the OP that I was referred to.

Notice that the expression of \(\displaystyle x-x^3+x^5\) forms a geometric progression with first term and common ratio as \(\displaystyle x\) and \(\displaystyle -x^2\) respectively. we see that the sum of the first three terms of this geometric series is:

\(\displaystyle S_3=\frac{a(1-r^3)}{1-r}=\frac{x(1+x^6)}{1+x^2}\)

Thus, f(x) can be rewritten as \(\displaystyle f(x)=-2+\frac{x(1+x^6)}{1+x^2}\)

Since \(\displaystyle f(\alpha)=0\), we have

\(\displaystyle f(\alpha)=-2+\frac{\alpha(1+\alpha^6)}{1+\alpha^2}=0\)

Rearrange the equation to make \(\displaystyle \alpha^6\) the subject, we obtain:

\(\displaystyle \alpha^6=-1+2\left(\frac{1+a^2}{a}\right)\)

AM-GM inequality tells us

\(\displaystyle \frac{a+\frac{1}{a}}{2}\ge \sqrt {a.\frac{1}{a}}\)

\(\displaystyle a+\frac{1}{a}\ge 2\) but we know in our case \(\displaystyle a\ne\frac{1}{a}\), the inequality that we have gotten become
\(\displaystyle a+\frac{1}{a}> 2\)

Thus,

\(\displaystyle \alpha^6>-1+2(2)>3\)

Now, we can safely say that \(\displaystyle [\alpha]=3\).
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
Re: Show that α^6=3.

I have found the solution suggested by someone who had solved this problem correctly and I want to share it with the members of MHB here.

We're given the equation of the function of f of x, i.e. \(\displaystyle f(x)=x^5-x^3+x-2\) and that \(\displaystyle \alpha\) is the real root of \(\displaystyle f(x)\).

First, we differentiate the function w.r.t.x and get:

\(\displaystyle f'(x)=5x^4-3x^2+1=5\left(\left(x^2-\frac{3}{10}\right)^2+\frac{11}{100}\right)\)

i.e. \(\displaystyle f'(x)\ge0\) for all real \(\displaystyle x\). This means \(\displaystyle f(x)\) is monotonically increasing, thus, \(\displaystyle f(x)\) has one and only one root, i.e. \(\displaystyle \alpha\).

From \(\displaystyle f(0)=-2\) and \(\displaystyle f'(2^{\frac{1}{3}})=2^{\frac{5}{3}}+2^{\frac{1}{3}}-4>0\), we can conclude by the Intermediate Value Theorem that the only root lies in \(\displaystyle (0, 2^{\frac{1}{3}})\).

Therefore, we get \(\displaystyle 0<\alpha<2^{\frac{1}{3}}\).

Raise both sides of the inequality to the sixth power we get:

\(\displaystyle 0<\alpha^6<4\)

Next, if we can prove that \(\displaystyle 3<\alpha^6<4\), we can also conclude that \(\displaystyle [\alpha]3\).

There maybe plenty of ways to prove that \(\displaystyle \alpha^6>3\) but this is what I did for I couldn't be able to fathom the working of the OP that I was referred to.

Notice that the expression of \(\displaystyle x-x^3+x^5\) forms a geometric progression with first term and common ratio as \(\displaystyle x\) and \(\displaystyle -x^2\) respectively. we see that the sum of the first three terms of this geometric series is:

\(\displaystyle S_3=\frac{a(1-r^3)}{1-r}=\frac{x(1+x^6)}{1+x^2}\)

Thus, f(x) can be rewritten as \(\displaystyle f(x)=-2+\frac{x(1+x^6)}{1+x^2}\)

Since \(\displaystyle f(\alpha)=0\), we have

\(\displaystyle f(\alpha)=-2+\frac{\alpha(1+\alpha^6)}{1+\alpha^2}=0\)

Rearrange the equation to make \(\displaystyle \alpha^6\) the subject, we obtain:

\(\displaystyle \alpha^6=-1+2\left(\frac{1+a^2}{a}\right)\)

AM-GM inequality tells us

\(\displaystyle \frac{a+\frac{1}{a}}{2}\ge \sqrt {a.\frac{1}{a}}\)

\(\displaystyle a+\frac{1}{a}\ge 2\) but we know in our case \(\displaystyle a\ne\frac{1}{a}\), the inequality that we have gotten become
\(\displaystyle a+\frac{1}{a}> 2\)

Thus,

\(\displaystyle \alpha^6>-1+2(2)>3\)

Now, we can safely say that \(\displaystyle [\alpha]=3\).
If that is the solution, then it seems to me this thread belongs in Calculus.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Show that α^6=3.

While the solution anemone found elsewhere utilizes the calculus, and her solution partially relies on it, I believe she is looking for a completely algebraic way to prove the statement. I think she just wanted to share a solution she found elsewhere after first posting the question.
 

Petrus

Well-known member
Feb 21, 2013
739
Re: Show that α^6=3.

While the solution anemone found elsewhere utilizes the calculus, and her solution partially relies on it, I believe she is looking for a completely algebraic way to prove the statement. I think she just wanted to share a solution she found elsewhere after first posting the question.
Well, the only agebra way I know is the way I posted. I did cheat with wolfram alpha that it only got one real root, that means we know \(\displaystyle \alpha\). Well I did try solve the equation but I got stuck so I wanted someone else to take over:)

Regards,
\(\displaystyle |\pi\rangle\)
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Re: Show that α^6=3.

I have found the solution suggested by someone who had solved this problem correctly and I want to share it with the members of MHB here.

We're given the equation of the function of f of x, i.e. \(\displaystyle f(x)=x^5-x^3+x-2\) and that \(\displaystyle \alpha\) is the real root of \(\displaystyle f(x)\).

First, we differentiate the function w.r.t.x and get:

\(\displaystyle f'(x)=5x^4-3x^2+1=5\left(\left(x^2-\frac{3}{10}\right)^2+\frac{11}{100}\right)\)

i.e. \(\displaystyle f'(x)\ge0\) for all real \(\displaystyle x\). This means \(\displaystyle f(x)\) is monotonically increasing, thus, \(\displaystyle f(x)\) has one and only one root, i.e. \(\displaystyle \alpha\).

From \(\displaystyle f(0)=-2\) and \(\displaystyle f'(2^{\frac{1}{3}})=2^{\frac{5}{3}}+2^{\frac{1}{3}}-4>0\), we can conclude by the Intermediate Value Theorem that the only root lies in \(\displaystyle (0, 2^{\frac{1}{3}})\).

Therefore, we get \(\displaystyle 0<\alpha<2^{\frac{1}{3}}\).

Raise both sides of the inequality to the sixth power we get:

\(\displaystyle 0<\alpha^6<4\)

Next, if we can prove that \(\displaystyle 3<\alpha^6<4\), we can also conclude that \(\displaystyle [\alpha]3\).

There maybe plenty of ways to prove that \(\displaystyle \alpha^6>3\) but this is what I did for I couldn't be able to fathom the working of the OP that I was referred to.

Notice that the expression of \(\displaystyle x-x^3+x^5\) forms a geometric progression with first term and common ratio as \(\displaystyle x\) and \(\displaystyle -x^2\) respectively. we see that the sum of the first three terms of this geometric series is:

\(\displaystyle S_3=\frac{a(1-r^3)}{1-r}=\frac{x(1+x^6)}{1+x^2}\)

Thus, f(x) can be rewritten as \(\displaystyle f(x)=-2+\frac{x(1+x^6)}{1+x^2}\)

Since \(\displaystyle f(\alpha)=0\), we have

\(\displaystyle f(\alpha)=-2+\frac{\alpha(1+\alpha^6)}{1+\alpha^2}=0\)

Rearrange the equation to make \(\displaystyle \alpha^6\) the subject, we obtain:

\(\displaystyle \alpha^6=-1+2\left(\frac{1+a^2}{a}\right)\)

AM-GM inequality tells us

\(\displaystyle \frac{a+\frac{1}{a}}{2}\ge \sqrt {a.\frac{1}{a}}\)

\(\displaystyle a+\frac{1}{a}\ge 2\) but we know in our case \(\displaystyle a\ne\frac{1}{a}\), the inequality that we have gotten become
\(\displaystyle a+\frac{1}{a}> 2\)

Thus,

\(\displaystyle \alpha^6>-1+2(2)>3\)

Now, we can safely say that \(\displaystyle [\alpha]=3\).
Hi anemone, :)

This solution would be much shorter (and sweeter) if you use the Intermediate value theorem considering the points, \(x=3^{1/6}\) and \(x=2^{1/3}\).
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683
Re: Show that α^6=3.

If that is the solution, then it seems to me this thread belongs in Calculus.
The only reason I posted this problem in Pre-Algebra forum was I didn't realize it required us to solve it using theorem of calculus and if I knew that was the case, I wouldn't post it in Pre-Algebra forum...sorry about that. Perhaps you wanted to move this to Calculus forum?

Hi anemone, :)

This solution would be much shorter (and sweeter) if you use the Intermediate value theorem considering the points, \(x=3^{1/6}\) and \(x=2^{1/3}\).
Thanks for sharing this useful hint with me...:eek:

Yes, you're right because \(\displaystyle f(x=3^{1/6})<0\) as well...

Thanks, Sudharaka!:)
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
Re: Show that α^6=3.

The only reason I posted this problem in Pre-Algebra forum was I didn't realize it required us to solve it using theorem of calculus and if I knew that was the case, I wouldn't post it in Pre-Algebra forum...sorry about that. Perhaps you wanted to move this to Calculus forum?
No issue. We don't infract if you don't know in advance that the solution involves higher math.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Re: Show that α^6=3.

The only reason I posted this problem in Pre-Algebra forum was I didn't realize it required us to solve it using theorem of calculus and if I knew that was the case, I wouldn't post it in Pre-Algebra forum...sorry about that. Perhaps you wanted to move this to Calculus forum?



Thanks for sharing this useful hint with me...:eek:

Yes, you're right because \(\displaystyle f(x=3^{1/6})<0\) as well...

Thanks, Sudharaka!:)
Hi anemone, :)

In addition to that you don't need to differentiate the polynomial and hence use calculus for this question. Instead the "not so famous" Budan's theorem comes in very handy in this situation. Using this we can show that there exist only one real root in the interval \((1,2)\) after which applying the Intermediate value theorem as usual will solve the problem.

Let, \(f(x)=x^5-x^3+x-2\)

\[f(x+1)={x}^{5}+5\,{x}^{4}+9\,{x}^{3}+7\,{x}^{2}+3x-1\]

Number of sign changes in the above equation is 1.

\[f(x+2)={x}^{5}+10\,{x}^{4}+39\,{x}^{3}+74\,{x}^{2}+69x+24\]

Number of sign changes in the above equation is 0.

So from Budan's theorem we have only one real root in the interval \((1,2)\). Now notice that both \(3^{1/6}\) and \(2^{1/3}\) lie in the interval \((1,2)\) and we know that there is only one root in this interval. Using the Intermediate value theorem as before will conclude the problem.
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683
Re: Show that α^6=3.

Hi anemone, :)

In addition to that you don't need to differentiate the polynomial and hence use calculus for this question. Instead the "not so famous" Budan's theorem comes in very handy in this situation. Using this we can show that there exist only one real root in the interval \((1,2)\) after which applying the Intermediate value theorem as usual will solve the problem.

Let, \(f(x)=x^5-x^3+x-2\)

\[f(x+1)={x}^{5}+5\,{x}^{4}+9\,{x}^{3}+7\,{x}^{2}+3x-1\]

Number of sign changes in the above equation is 1.

\[f(x+2)={x}^{5}+10\,{x}^{4}+39\,{x}^{3}+74\,{x}^{2}+69x+24\]

Number of sign changes in the above equation is 0.

So from Budan's theorem we have only one real root in the interval \((1,2)\). Now notice that both \(3^{1/6}\) and \(2^{1/3}\) lie in the interval \((1,2)\) and we know that there is only one root in this interval. Using the Intermediate value theorem as before will conclude the problem.
Hi Sudharaka, thank you for sharing with me of this Budan's theorem and showing me how to apply it to this problem......this is the second time you introduced some very useful theorems or identities to me and I am very grateful for this.:)
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Re: Show that α^6=3.

Hi Sudharaka, thank you for sharing with me of this Budan's theorem and showing me how to apply it to this problem......this is the second time you introduced some very useful theorems or identities to me and I am very grateful for this.:)
My pleasure. :)
 

zzephod

Well-known member
Feb 3, 2013
134
Re: Show that α^6=3.

I'm very glad to be able to finally locate the original problem and frankly, I still need help on this problem...:eek:

Problem:
If \(\displaystyle \alpha\) is a real root of \(\displaystyle f(x)=x^5-x^3+x-2\), then prove that \(\displaystyle [\alpha^6]=3\) where \(\displaystyle [a]\) denotes the greatest integer function.
This is equivalent to showing that the root (once you are satisfied that there is but one root) is in \(\displaystyle \left[3^{1/6}, 4^{1/6}\right)\), which can be shown by seeing that \(\displaystyle f(x)\) changes sign between \(\displaystyle x=1.201>3^{1/6}\) and \(\displaystyle x=1.25<4^{1/6}\).

.