MHB Zeros of reciprocal functions and transformed

[Amad27]

Hello,

I was recently browsing the thread,

http://mathhelpboards.com/challenge-questions-puzzles-28/evaluate-1-1-1-b-1-b-1-c-1-c-12074.html

And I was browsing both, Kaliprasad's method and anemone's method.

Lets take Kaliprasad's approach first. My question is about the reciprocal zeros, NOT the actual problem... =)

Anywho, it says at the start of his answer,

$a, b, c$ are roots of $x^3 - x - 1 = 0$ and therefore,
$\frac{1}{a} , \frac{1}{b}, \frac{1}{c}$ are the roots of $\frac{1}{x^3} - \frac{1}{x} - 1$

I just want to know the mechanism, behind this. How is this true, what makes this true (more or less)? SECONDLY, anemone's answer, she states,

"We're told that $a, b, c$ are the roots of $f(x) = x^3 - x - 1$, hence the function,

$f(x - 1) = (x-1)^3 - (x-1) - 1 = x^3 - 3x^2 + 2x - 1$ has roots $(a+1), (b+1), (c+1)$

Again, what is the mechanism?

Is there are proof for any of the two?

THANKS!

 

[Deveno]

Hello,

I was recently browsing the thread,

http://mathhelpboards.com/challenge-questions-puzzles-28/evaluate-1-1-1-b-1-b-1-c-1-c-12074.html

And I was browsing both, Kaliprasad's method and anemone's method.

Lets take Kaliprasad's approach first. My question is about the reciprocal zeros, NOT the actual problem... =)

Anywho, it says at the start of his answer,

$a, b, c$ are roots of $x^3 - x - 1 = 0$ and therefore,
$\frac{1}{a} , \frac{1}{b}, \frac{1}{c}$ are the roots of $\frac{1}{x^3} - \frac{1}{x} - 1$

I just want to know the mechanism, behind this. How is this true, what makes this true (more or less)? SECONDLY, anemone's answer, she states,

"We're told that $a, b, c$ are the roots of $f(x) = x^3 - x - 1$, hence the function,

$f(x - 1) = (x-1)^3 - (x-1) - 1 = x^3 - 3x^2 + 2x - 1$ has roots $(a+1), (b+1), (c+1)$

Again, what is the mechanism?

Is there are proof for any of the two?

THANKS!

Suppose that we have:

$f(x) = x^3 - x - 1 = (x - a)(x - b)(x - c) = 0$.

Suppose further, that we let $y = \dfrac{1}{x}$ (assuming that none of $a,b,c$ are 0).

Then $x = \dfrac{1}{y}$ so:

$f(x) = f\left(\frac{1}{y}\right) = \left(\frac{1}{y} - a\right)\left(\frac{1}{y} - b\right)\left(\frac{1}{y} - c\right)$.

If this is 0, one of the three factors must be 0, so, for example, if:

$\frac{1}{y} - a = 0$ then:

$\frac{1}{y} = a$, and thus:

$y = \frac{1}{\left(\frac{1}{y}\right)} = \frac{1}{a}$.

This is because if $c,d \neq 0$ then:

$c = d \iff \frac{1}{c} = \frac{1}{d}$.

Remember we decide equality of two fractions by "cross-multiplying":

if $\frac{1}{c} = \frac{1}{d}$, this means that $(1)(d) = (c)(1)$, that is $c = d$.

Now there is one thing we need to be careful of: we need to be sure 0 is not a root of $f$.

But $f$ has a non-zero constant term, that is: $f(0) \neq 0$, so we are good.

See if you can do the same thing with your second question, letting $y = x + 1$.

 

[Amad27]

Suppose that we have:

$f(x) = x^3 - x - 1 = (x - a)(x - b)(x - c) = 0$.

Suppose further, that we let $y = \dfrac{1}{x}$ (assuming that none of $a,b,c$ are 0).

Then $x = \dfrac{1}{y}$ so:

$f(x) = f\left(\frac{1}{y}\right) = \left(\frac{1}{y} - a\right)\left(\frac{1}{y} - b\right)\left(\frac{1}{y} - c\right)$.

If this is 0, one of the three factors must be 0, so, for example, if:

$\frac{1}{y} - a = 0$ then:

$\frac{1}{y} = a$, and thus:

$y = \frac{1}{\left(\frac{1}{y}\right)} = \frac{1}{a}$.

This is because if $c,d \neq 0$ then:

$c = d \iff \frac{1}{c} = \frac{1}{d}$.

Remember we decide equality of two fractions by "cross-multiplying":

if $\frac{1}{c} = \frac{1}{d}$, this means that $(1)(d) = (c)(1)$, that is $c = d$.

Now there is one thing we need to be careful of: we need to be sure 0 is not a root of $f$.

But $f$ has a non-zero constant term, that is: $f(0) \neq 0$, so we are good.

See if you can do the same thing with your second question, letting $y = x + 1$.

Hi, just a remark,

$f(x) = f\left(\frac{1}{y}\right) = \left(\frac{1}{y} - a\right)\left(\frac{1}{y} - b\right)\left(\frac{1}{y} - c\right)$.

Then you get (properly),

$\frac{1}{y} - a = 0$
Which becomes

$\frac{1}{y} = a$
Finally,

$y = \frac{1}{\left(\frac{1}{y}\right)} = \frac{1}{a}$

This proves the zero is at $y = \frac{1}{a}$

We were wondering how it proves the zero is at $x = \frac{1}{a}$??

 

[Amad27]

Suppose that we have:

$f(x) = x^3 - x - 1 = (x - a)(x - b)(x - c) = 0$.

Suppose further, that we let $y = \dfrac{1}{x}$ (assuming that none of $a,b,c$ are 0).

Then $x = \dfrac{1}{y}$ so:

$f(x) = f\left(\frac{1}{y}\right) = \left(\frac{1}{y} - a\right)\left(\frac{1}{y} - b\right)\left(\frac{1}{y} - c\right)$.

If this is 0, one of the three factors must be 0, so, for example, if:

$\frac{1}{y} - a = 0$ then:

$\frac{1}{y} = a$, and thus:

$y = \frac{1}{\left(\frac{1}{y}\right)} = \frac{1}{a}$.

This is because if $c,d \neq 0$ then:

$c = d \iff \frac{1}{c} = \frac{1}{d}$.

Remember we decide equality of two fractions by "cross-multiplying":

if $\frac{1}{c} = \frac{1}{d}$, this means that $(1)(d) = (c)(1)$, that is $c = d$.

Now there is one thing we need to be careful of: we need to be sure 0 is not a root of $f$.

But $f$ has a non-zero constant term, that is: $f(0) \neq 0$, so we are good.

See if you can do the same thing with your second question, letting $y = x + 1$.

I reflected upon this, and came up with a new, better idea.

$f(x) = x^3 - x + 1$

$f(\frac{1}{x}) = \frac{1}{x^3} - \frac{1}{x} + 1 = (\frac{1}{x}-a)(\frac{1}{x}-b)(\frac{1}{x}-c)$Considering $a, b, c$ being the roots.

Let's set $f(\frac{1}{x}) = 0$ This would imply,

$(\frac{1}{x}-a)(\frac{1}{x}-b)(\frac{1}{x}-c) = 0$

Which breaks up into three parts,

$(\frac{1}{x}-a) = 0 \implies x = \frac{1}{a}$
$(\frac{1}{x}-b) = 0 \implies x = \frac{1}{b}$
$(\frac{1}{x}-c) = 0 \implies x = \frac{1}{c}$

We have proved that the reciprocal,

$f(x) \implies f(\frac{1}{x})$ has zeros, which are reciprocals themselves, since.

$f(\frac{1}{x}) = \frac{1}{f(x)}$

Thanks! (There is no question involved, I just want to put this out there, if anyone else was confused!)

 

[Evgeny.Makarov]

Let's set $f(\frac{1}{x}) = 0$ This would imply,

$(\frac{1}{x}-a)(\frac{1}{x}-b)(\frac{1}{x}-c) = 0$

Which breaks up into three parts,

$(\frac{1}{x}-a) = 0 \implies x = \frac{1}{a}$
$(\frac{1}{x}-b) = 0 \implies x = \frac{1}{b}$
$(\frac{1}{x}-c) = 0 \implies x = \frac{1}{c}$

We have proved that the reciprocal,

Up to here it makes sense.

$f(x) \implies f(\frac{1}{x})$ has zeros, which are reciprocals themselves

I would say, $x_0$ is a root of $f(x)$ $\implies$ $1/x_0$ is a root of $f(1/x)$ (provided $x_0\ne0$).

since.

$f(\frac{1}{x}) = \frac{1}{f(x)}$

But this is definitely not true, even for this $f(x)$.

$a, b, c$ are roots of $x^3 - x - 1 = 0$ and therefore,
$\frac{1}{a} , \frac{1}{b}, \frac{1}{c}$ are the roots of $\frac{1}{x^3} - \frac{1}{x} - 1$

...

We're told that $a, b, c$ are the roots of $f(x) = x^3 - x - 1$, hence the function,

$f(x - 1) = (x-1)^3 - (x-1) - 1 = x^3 - 3x^2 + 2x - 1$ has roots $(a+1), (b+1), (c+1)$

Again, what is the mechanism?

In general, suppose that $f(x_0)=y_0$ and let $F(x)=f(g(x))$. Let $x_1=g^{-1}(x_0)$, i.e., $g(x_1)=x_0$, if such $x_1$ exists. Then $F(x_1)=y_0$. Indeed, $F(x_1)=f(g(x_1))=f(x_0)=y_0$. What you are doing in forming $F$ is pre-applying $g$ to the argument before applying $f$. But $x_1$ cancels this because it applies the inverse of $g$ to $x_0$. We have the following chain.
\[
x_0\overset{g^{-1}}{\longmapsto} x_1\mathrel{\overbrace{\overset{g}{\longmapsto} x_0\overset{f}{\longmapsto}}^F} y_0
\]
In Kaliprasad's approach, $g(x)=1/x$ and $x_1=1/x_0$, so that $g(x_1)=x_0$. In anemone's approach, $g(x)=x-1$ and $x_1=x_0+1$, so that $g(x_1)=x_0$.

 

[Amad27]

Up to here it makes sense.

I would say, $x_0$ is a root of $f(x)$ $\implies$ $1/x_0$ is a root of $f(1/x)$ (provided $x_0\ne0$).

But this is definitely not true, even for this $f(x)$.

In general, suppose that $f(x_0)=y_0$ and let $F(x)=f(g(x))$. Let $x_1=g^{-1}(x_0)$, i.e., $g(x_1)=x_0$, if such $x_1$ exists. Then $F(x_1)=y_0$. Indeed, $F(x_1)=f(g(x_1))=f(x_0)=y_0$. What you are doing in forming $F$ is pre-applying $g$ to the argument before applying $f$. But $x_1$ cancels this because it applies the inverse of $g$ to $x_0$. We have the following chain.
\[
x_0\overset{g^{-1}}{\longmapsto} x_1\mathrel{\overbrace{\overset{g}{\longmapsto} x_0\overset{f}{\longmapsto}}^F} y_0
\]
In Kaliprasad's approach, $g(x)=1/x$ and $x_1=1/x_0$, so that $g(x_1)=x_0$. In anemone's approach, $g(x)=x-1$ and $x_1=x_0+1$, so that $g(x_1)=x_0$.

Yeah, just realized, but I hope you knew what I meant.

I did not means

$f(\frac{1}{x}) = \frac{1}{f(x)}$, I meant this for the roots! Sorry =)

We could take a similar approach to anemone's approach.

$f(x) = x^3 - x - 1$

$f(x) = (x - a)(x - b)(x - c) $

$f(x - 1) = (h - a)(h - b)(h - c)$ where $h = x - 1$

Now here we also have three simple equations,

$x = a + 1$
$x = b + 1$
$x = c + 1$

Thanks!