Order the given radical numbers

[littlemathquark]

TL;DR Summary: $x=\sqrt[7]{13}+\sqrt[6]{13}$, $y=\sqrt[5]{13}+\sqrt[8]{13}$ , $z=\sqrt[3]{13}+\sqrt[10]{13}$

$x=\sqrt[7]{13}+\sqrt[6]{13}$ , $y=\sqrt[5]{13}+\sqrt[8]{13}$ , $z=\sqrt[3]{13}+\sqrt[10]{13}$

Order the given radical numbers.

I have no idea. I don't want to use calculator. May be I can use arithmetic mean, geometric-mean but I'm not sure. Any clue?

 

[jbriggs444]

$x=\sqrt[7]{13}+\sqrt[6]{13}$ , $y=\sqrt[5]{13}+\sqrt[8]{13}$ , $z=\sqrt[3]{13}+\sqrt[10]{13}$

May be I can use arithmetic mean, geometric-mean but I'm not sure.

Drop the idea of a mean. That is not the intended approach.

You could make the observation that the sum of the orders in each of the three expressions is the same constant. And that the base under the radical sign is the same for all of the terms.

 

[littlemathquark]

Yes, it's 13 but I could'nt understand how I use that fact still.

 

[fresh_42]

Yes, it's 13 but I could'nt understand how I use that fact still.

Which methods are you allowed to use? E.g. is differentiation allowed?

 

[littlemathquark]

İt's open all methods.

 

[fresh_42]

Then you have to compare the expressions
$$
f(x)=13^{1/x} + 13^{1/(13-x)}
$$
for different values of $x>0.$ It has a local minimum which could help.

 

[fresh_42]

I haven't checked, but if you want to use means, then harmonic < geometric < arithmetic could help. I'm not sure though.

 

[jbriggs444]

Even without differentiation, one can sketch the curve and see that $f(x) = 13^{1/x}$ is concave upward. It follows that the mirror image function $f(x) = 13^{1/(13-x)}$ is also concave upward.

So the sum, $f(x) = 13^{1/x} + 13^{1/(13-x)}$ must be concave upward. That function has an obvious line of symmetry at $x=6.5$. It follows that it has its global minimum at $x=6.5$ and is strictly monotone increasing for $x > 6.5$

 

[littlemathquark]

I haven't checked, but if you want to use means, then harmonic < geometric < arithmetic could help. I'm not sure though.

Thanks, I will try.

 

[fresh_42]

It follows that it has its global minimum at $x=6.5$ and is strictly monotone increasing for $x > 6.5$

Sure?

That's where I gave up my considerations since we only need values $0<x< 6.5$
But $f(6.5)\approx 3$ and $\displaystyle{\lim_{x \to \infty}f(x)=2.}$

 

[jbriggs444]

Sure?

That's where I gave up my considerations since we only need values $0<x< 6.5$
But $f(6.5)\approx 3$ and $\displaystyle{\lim_{x \to \infty}f(x)=2.}$

So you are concentrating on the left hand side where $0 \lt x \lt 6.5$. Fair enough.

Can you see that the function in question is strictly monotone decreasing in that range?

 

[fresh_42]

So you are concentrating on the left hand side where $0 \lt x \lt 6.5$. Fair enough.

Can you see that the function in question is strictly monotone decreasing in that range?

Yes, that's the easy part and sufficient for the problem. What I have no idea about is how it looks for $x>6.5$ since it has to come from a minimum $(6.5,3)$ to approach $(\infty ,2)$ and looks concave. That was a bit strange and I didn't resolve the apparent contradiction as it was irrelevant for the problem.

 

[jbriggs444]

Yes, that's the easy part and sufficient for the problem. What I have no idea about is how it looks for $x>6.5$ since it has to come from a minimum $(6.5,3)$ to approach $(\infty ,2)$ and looks concave. That was a bit strange and I didn't resolve the apparent contradiction as it was irrelevant for the problem.

We should probably restrict our attention to the open interval $(0,13)$ where the function is well behaved. The function is nicely symmetric within this interval. The approach on the right hand side is to $(13,\infty)$.

We have poles at $x=0$ and $x=13$ which are to be avoided.

 

[fresh_42]

We should probably restrict our attention to the open interval (0,13) where the function is well behaved. The function is nicely symmetric within this interval. The approach on the right hand side is to $(13,13^{13}+1)$.

Yes. However, it caused me to correct my post from global to local minimum since I didn't restrict the domain in the first place. That was basically where I wasn't completely satisfied by your post #8 since you didn't restrict it either.

 

[jbriggs444]

Yes. However, it caused me to correct my post from global to local minimum since I didn't restrict the domain in the first place. That was basically where I wasn't completely satisfied by your post #8 since you didn't restrict it either.

I thought about it, but did not want to add clutter. My bad.

 

[littlemathquark]

Thank you for your helps. Short solution is here:

$x=f(7)$, $y=f(8)$, $z=f(10)$ and $f$ is increasing function for $>6,5$ then $z>y>x$

 

[littlemathquark]

İnfact I observed that for example let $a=\sqrt[2]{64}+\sqrt[4]{64}=8+2\sqrt 2$ and $b=\sqrt[3]{64}+\sqrt[3]{64}=8$ and $a>b$ , I mean as the differences between the degrees of the roots increase, the total becomes larger. Because of this $z>y>x$. Is it true every time?

 

[littlemathquark]

I think we have another solution using mean value theorem.

 

[littlemathquark]

I haven't checked, but if you want to use means, then harmonic < geometric < arithmetic could help. I'm not sure though.

I think it can be given another solution using mean value theorem but I have no idea about using a-g-h mean. Could they help?

 

[fresh_42]

I think it can be given another solution using mean value theorem but I have no idea about using a-g-h mean. Could they help?

I doubt it. It was your idea. I don't even see how the MVT could help. It proves the existence of a point somewhere in between, but we have three given points that must be compared. How would another unknown point help?

 

[littlemathquark]

Like this :
Let $f(t) =13^{1/t}$
As $f''(t) >0$ then $f'$ is increasing function. So $f'(5)<(f(6)-f(5))/(6-5)<f'(6)$
Similarly $f'(7)<(f(8)-f(7))/(8-7)<f'(8)$
thus we are able to show $x<y$

 

[fresh_42]

Like this :
Let $f(t) =13^{1/t}$
As $f''(t) >0$ then $f'$ is increasing function. So $f'(5)<(f(6)-f(5))/(6-5)<f'(6)$
Similarly $f'(7)<(f(8)-f(7))/(8-7)<f'(8)$
thus we are able to show $x<y$

This isn't the MVT. It is simply the fact that $f(x)=13^{1/x}+13^{1/(13-x)}$ decreases for $0<x<6.5$ and increases for $6.5< x< 13.$

 

[littlemathquark]

Yes, you are right.; my mistake.
$f(x)=13^{1/x}$, $f^\prime (x) = -\dfrac{13^{1/x}}{x^2}$

When $x>0$, $f(x)$ is increasing, $f^\prime (x) <0$ and $|f^\prime (x)|$ is increasing. It shows that the rate of increase of $f$ decreases as $x$ increases.

So $f(3)-f(5) >f( 8 )-f(10)$ and $f(5)-f(6)>f(7)-f( 8 )$, thus $z>y>x$

Is it possible to solve the problem using means?

 

[fresh_42]

Yes, you are right.; my mistake.
$f(x)=13^{1/x}$, $f^\prime (x) = -\dfrac{13^{1/x}}{x^2}$

When $x>0$, $f(x)$ is increasing, $f^\prime (x) <0$ and $|f^\prime (x)|$ is increasing. It shows that the rate of increase of $f$ decreases as $x$ increases.

You have to consider the function $f(x)=13^{1/x}+13^{1/(13-x)}$ not only the part that is easier to differentiate. How else could you know that the second term does not change the behavior of the first one? Otherwise, you have two variables, not one.

So $f(3)-f(5) >f( 8 )-f(10)$ and $f(5)-f(6)>f(7)-f( 8 )$, thus $z>y>x$

That does not convince me with your choice of $f(x).$ However, I did not try to complete all the arguments you left out. Too handwavy for my taste.

Is it possible to solve the problem using means?

No, not that I found a way.

 

[littlemathquark]

Let's define the function $f(t) = 13^{\frac{1}{t}} + 13^{\frac{1}{13-t}}$ on the interval $t \in (0, 13)$. If we write $x = f(6)$, $y = f(5)$, and $z = f(3)$, we can make a prediction. Since $f(13-t) = f(t)$, the function is symmetric about $t = \frac{13}{2} = 6.5$. At $t = 0$, or rather in the limit, we have$f(0^+) = 13^{+\infty} + 13^{\frac{1}{13}} = \infty$so we can "predict" that the function is decreasing on the interval $(0, 6.5)$ and increasing on the interval $(6.5, 13)$.
According to this prediction, we will have $z > y > x$.To prove the prediction, it is sufficient to show that the derivative of the function is $0$ only at $t = 6.5$. We obtain$f'(t) = \ln 13 \left[ \frac{13^{\frac{1}{13-t}}}{(13-t)^2} - \frac{13^{\frac{1}{t}}}{t^2} \right]$It can be seen that $f'(t) = 0$ for $t = 6.5$. If $t > 6.5$, then$t^2 > (13-t)^2 \text{ and } 13^{\frac{1}{13-t}} > 13^{\frac{1}{t}} \implies f'(t) > 0$and similarly, if $t < 6.5$, then $f'(t) < 0$. This completes the proof.

 

[littlemathquark]

İs it possible to solve the question without using differentation?

 

[fresh_42]

İs it possible to solve the question without using differentation?

I haven't found one. The basic difficulty is that addition and multiplication do not match well. We can rewrite the problem as: Compare the function values of the function $g(x)=13^{6.5 - x}+13^{6.5 + x}\, , \,0<x<6.5$
at $x=3.5\, , \,1.5\, , \,0.5.$ If we look at its graph

then we see that it is relatively flat and only slightly increasing for $0<x\leq 3.5.$ This isn't a proof, but it demonstrates that we do not have much room for rough estimations. I would guess that even series expansions have to be considered up to several terms. Maybe you can show that it is convex, but I doubt that would be easier since the basic problem with the addition gets even worse.

 

[littlemathquark]

I haven't found one. The basic difficulty is that addition and multiplication do not match well. We can rewrite the problem as: Compare the function values of the function $g(x)=13^{6.5 - x}+13^{6.5 + x}\, , \,0<x<6.5$
at $x=3.5\, , \,1.5\, , \,0.5.$ If we look at its graph

View attachment 357763

then we see that it is relatively flat and only slightly increasing for $0<x\leq 3.5.$ This isn't a proof, but it demonstrates that we do not have much room for rough estimations. I would guess that even series expansions have to be considered up to several terms. Maybe you can show that it is convex, but I doubt that would be easier since the basic problem with the addition gets even worse.

Could it be like this? Since 13^x is increasing, 13^(1/x) = f(x) is decreasing. As X increases without bound, as the f function approaches 1, the difference between the x's approaches 0.

When 3 < 5 < 6 < 7 < 8 < 10, then f(3) - f(5) < f(8) - f(10) and f(5) - f(6) < f(7) - f(8).

 

[fresh_42]

Could it be like this? Since 13^x is increasing, 13^(1/x) = f(x) is decreasing. As X increases without bound, as the f function approaches 1, the difference between the x's approaches 0.

When 3 < 5 < 6 < 7 < 8 < 10, then f(3) - f(5) < f(8) - f(10) and f(5) - f(6) < f(7) - f(8).

See my post #24. I am not convinced. You have no quantitative information about the differences. Your argument seems only to mimic the differentiation argument in a handwavy way.

How do you formally get from $f(3) > f(5) > f(6) > f(7)> f(8)> f(10)$ to $f(3)-f(5)>f(8)-f(10)\,?$ They are both positive, but how do we know that the right one is less? You factually make a statement about the second derivative, the behavior of the difference quotients.

Sure, you can always mimic differentiation but why? This is only more troublesome when making it rigorous and more lengthy. If we take both terms into account then we will need only the first derivative on half of the domain.

 

[pasmith]

Set x= \frac{13 + u}2. Then \begin{split}<br /> 13^{1/x} + 13^{1/(13-x)} &amp;= 13^{2/(13 + u)} + 13^{2/(13 - u)} \\<br /> &amp;= 169^{(13-u)/(169-u^2)} + 169^{(13 + u)/(169 - u^2)} \\<br /> &amp;= 2\cdot 169^{13/(169-u^2)}\cosh\left( \frac{u \ln 169}{169 - u^2}\right).\end{split} For 0 \leq u &lt; 13 both factors are strictly increasing.

 

[mathwonk]

how's this?
A = 1/5 - 1/7 > 1/6 - 1/8 = a, hence (13)^(1/5) - (13)^(1/7)

=[(13)^(1/7][13^A - 1] > [(13)^(1/8)][13^a - 1]

= (13)^(1/6) - (13)^(1/8), hence

y = (13)^(1/5) + (13)^(1/8) > (13)^(1/6) + (13)^(1/7) = x.

just elementary precalculus, unless I made a mistake, which I frequently do in this realm.

edit: I guess this is basically the same as pasmith's (more general) solution in the previous post.