Minimize sum of number & reciprical + proof

[Dethrone]

Find a positive number such that the sum of the number and its reciprocal is as small as possible. Full marks for proving your answer is correct.

Process:

let $a>0$
$$f(a)=a+\frac{1}{a}$$
$$f'(a)=1-\frac{1}{a^2}=\frac{(a+1)(a-1)}{a^2}$$

The critical numbers are $0$, $\pm 1$, but only $1$ is a solution because it is greater than 0. Now I have to prove this, and I'm pretty sure it's not by doing a first or second derivative test. Now I'm kind of excited right now because it's my first time doing a "proof by contradiction". Let me know if it works :D

Suppose there is a $b>0$ such that its sum and reciprocal is less than $a$.
$$b+\frac{1}{b}<a+\frac{1}{a}$$
The RH side is equal to $2$:
$$b+\frac{1}{b}<2$$
$$b^2-2b+1<0$$
$$(b-1)^2<0$$

There are no solutions that satisfy the inequality. Hence, $a=1$ is the smallest number that satisfy the problem. Let me know whether my proof works or not, or if there are better ways of proving it :D

 

[MarkFL]

I don't see anything wrong with your proof, however you could also use either the first or second derivative tests, or you could take the limit of the objective function at the boundaries, which are $0^{+}$ and $\infty$ and draw a conclusion from that.

 

[Dethrone]

The limit as you approach infinity and $0^+$of the objective function is infinity. How can we draw a conclusion from that?

 

[MarkFL]

The limit as you approach infinity and $0^+$of the objective function is infinity. How can we draw a conclusion from that?

Well, if the objective function is unbounded at the end-points, and only has 1 critical number within the interval for which the objective function has a finite value, then what must be the nature of the extremum associated with this critical number?

To be honest though, I would use the second derivative test, as it is easy to use here. :D

 

[Euge]

Find a positive number such that the sum of the number and its reciprocal is as small as possible. Full marks for proving your answer is correct.

Process:

let $a>0$
$$f(a)=a+\frac{1}{a}$$
$$f'(a)=1-\frac{1}{a^2}=\frac{(a+1)(a-1)}{a^2}$$

The critical numbers are $0$, $\pm 1$, but only $1$ is a solution because it is greater than 0. Now I have to prove this, and I'm pretty sure it's not by doing a first or second derivative test. Now I'm kind of excited right now because it's my first time doing a "proof by contradiction". Let me know if it works :D

Suppose there is a $b>0$ such that its sum and reciprocal is less than $a$.
$$b+\frac{1}{b}<a+\frac{1}{a}$$
The RH side is equal to $2$:
$$b+\frac{1}{b}<2$$
$$b^2-2b+1<0$$
$$(b-1)^2<0$$

There are no solutions that satisfy the inequality. Hence, $a=1$ is the smallest number that satisfy the problem. Let me know whether my proof works or not, or if there are better ways of proving it :D

Hi Rido12,

Note that for $a > 0$,

$\displaystyle f(a) = a - 2 + \frac{1}{a} + 2 = \left(\sqrt{a} - \frac{1}{\sqrt{a}}\right)^2 + 2 \ge 2$,

with equality if and only if $(\sqrt{a} - \frac{1}{\sqrt{a}})^2 = 0$, i.e., $\sqrt{a} = \frac{1}{\sqrt{a}}$. The solution to $\sqrt{a} = \frac{1}{\sqrt{a}}$ is $a = 1$. Hence $f(a)$ is minimized at $a = 1$ with minimum value 2.

 

[Dethrone]

Hi Euge (Wave)!

That's a really great way of looking at that problem...I would have never thought about it. A way of algebraically optimizing a function (Cool)

Mark,

I think I jumped to conclusions too fast and assumed that there was no way that the answer needed only the 2nd derivative test... My conclusion was right, but I jumped too fast. (Giggle) Because for part 2 of the question, I don't think the 2nd derivative test will work.

Question: Find all numbers such that the absolute value of the difference between the number and its reciprocal is as small as possible. Full marks for proving your answer is correct.

$$f(x)=\left| a-\frac{1}{a} \right|=\left| \frac{a^2-1}{a} \right|=\begin{cases}\frac{a^2-1}{a}, & -1<a<0, a>1 \\[3pt] -\frac{a^2-1}{a}, & a<-1 , 0<a<1 \\ \end{cases}$$

Differentiating each case, I get:
$$f'(x)=\frac{a^2+1}{a^2} \text{ or} -\frac{a^2+1}{a^2}$$

Critical number is $0$, which I suspect is the minimum of $f(x)$, but putting it in for the second derivative test gives me undefined. So now I must use a proof by contradiction, or one of Euge's ingenious proofs?

 

[Deveno]

Hi Euge (Wave)!

That's a really great way of looking at that problem...I would have never thought about it. A way of algebraically optimizing a function (Cool)

Mark,

I think I jumped to conclusions too fast and assumed that there was no way that the answer needed only the 2nd derivative test... My conclusion was right, but I jumped too fast. (Giggle) Because for part 2 of the question, I don't think the 2nd derivative test will work.

Question: Find all numbers such that the absolute value of the difference between the number and its reciprocal is as small as possible. Full marks for proving your answer is correct.

$$f(x)=\left| a-\frac{1}{a} \right|=\left| \frac{a^2-1}{a} \right|=\begin{cases}\frac{a^2-1}{a}, & -1<a<0, a>1 \\[3pt] -\frac{a^2-1}{a}, & a<-1 , 0<a<1 \\ \end{cases}$$

Differentiating each case, I get:
$$f'(x)=\frac{a^2+1}{a^2} \text{ or} -\frac{a^2+1}{a^2}$$

Critical number is $0$...

I have no idea what you mean by this remark, the extrema are $a = \pm 1$

... which I suspect is the minimum of $f(x)$, but putting it in for the second derivative test gives me undefined. So now I must use a proof by contradiction, or one of Euge's ingenious proofs?

An absolute value is minimum at 0 (0 is the smallest non-negative real number) so it suffices to find all $a$ for which:

$a = \dfrac{1}{a}$.

Clearly, 1 and -1 work. Can there be any others?

If $a > 1$ then $\dfrac{1}{a} < 1$, so that:

$a - \dfrac{1}{a} > 1 - \dfrac{1}{a} > 1 - 1 = 0$

If $0 < a < 1$, then $\dfrac{1}{a} > 1$, so that:

$\left|a - \dfrac{1}{a}\right| = \dfrac{1}{a} - a > 1 - a > 1 - 1 = 0$

The other cases can be handled similarly. In all cases, we find $a \neq \dfrac{1}{a}$ so their difference is going to be non-zero.

I think the point of this, is to show that "size" (absolute value of a number) is, in some respects easier to handle than "value", because with size we have an "absolute" hard boundary at the bottom. Compare this with finding extrema of a function which may vary between $-\infty$ and $\infty$, for which we might, at best, only find a "locally optimum" solution.

 

[Dethrone]

Before I proceed reading, why are the critical numbers $\pm 1$?

We have:

$$f'(x)=\frac{a^2+1}{a^2} \text{ or} -\frac{a^2+1}{a^2}$$

Setting $f'(x)=0 \text{ or D.N.E}$, we obtain that $a=0$
Or am I not allowed to take the derivative that way? (Wondering)

 

[Deveno]

Before I proceed reading, why are the critical numbers $\pm 1$?

We have:

$$f'(x)=\frac{a^2+1}{a^2} \text{ or} -\frac{a^2+1}{a^2}$$

Setting $f'(x)=0 \text{ or D.N.E}$, we obtain that $a=0$

Did you check the points at which $f$ is not differentiable?

$f$ is undefined at 0, so it's not even in the domain.

 

[Dethrone]

I guess I calculated the wrong derivative...because we're not allowed to differentiate the cases separately? I've always done that for $\d{}{x}\left| x \right|$

So do I rewrite the function as this and differentiate?:

$$f(x)=\left| a-\frac{1}{a} \right|=\sqrt{\left(a-\frac{1}{a}\right)^2}$$

EDIT: I have to go soon, but I'll be back tonight :D

 

[Deveno]

You can't find the minimum of $f(x) = |x|$ by differentiating, it doesn't fulfill the pre-conditions of the requisite theorems (it's not differentiable everywhere in any neighborhood of its minimum).

If we differentiate $f(x) = |x|$ where we can, we get:

$f'(x) = 1, x > 0$

$f'(x) = -1, x < 0$.

This tells us nothing about what happens AT 0.

Differentiating isn't a "cure-all". You have to verify differentiability FIRST, which is a rather stringent condition.

 

[Euge]

Hello again Rido12,

To answer your second question, perhaps it'll be easier to find a solution by considering $f(a)^2$, whether you use calculus or not. Without calculus, you can solve the problem by noting that $f(a)^2 \ge 0$, with equality if and only if $a - 1/a = 0$. The solutions to $a - 1/a = 0$ are $a = \pm 1$. Hence $f(a)$ is minimized at $a = 1$ and $a = -1$, with minimum value 0.

If you must use calculus, then take the derivative of $f(a)^2$ and set it equal to 0:

$\displaystyle (f^2)'(a) = 2\left(a - \frac{1}{a}\right) \left(1+ \frac{1}{a^2}\right) = 0$.

Since $1 + \frac{1}{a^2} > 0$, we have $a - \frac{1}{a} = 0$, so $a = \pm 1$. So $f^2$ has critical points at $a = 1$ and $a = -1$. Now

$\displaystyle (f^2)''(a) = 2\left(1 + \frac{3}{a^4}\right) > 0$

Hence, by the second derivative test, $f^2$ has a relative minimum at $a = -1$ and $a = 1$. It follows that $f^2$ is minimized at these points. So $f$ is also minimized at these points.

 

[Dethrone]

Okay, I understand now. The absolute value function is not differentiable at $x=0$, so differentiating it will tell us nothing about $x=0$. So we can either set the function equal to zero, which is the minimum of the function, or we can square the function (effectively eliminating the absolute value) and differentiate. Am I right?

Also, I know that an absolute value function must always be greater than zero, but must it always be greater and equal to zero? If an absolute value function does not equal zero, than it seems that we can only use the calculus method of solving...

 

[Euge]

You don't set the function equal to 0. You prove that $f(a)$ has minimum value 0 at $a = 1$ and $a = -1$.

By definition of absolute value, the absolute value of a real number is nonnegative. Certainly $|0| = 0$, so the absolute value function cannot be strictly positive. In fact, $|x| = 0$ if and only if $x = 0$.

 

[Dethrone]

Yes, I was referring to your calculus method where you found $f^2(a)$ and solved for its critical numbers by differentiating it. I think I understand this now. Although, I prefer both of your non-calculus methods better.