Cubic positive solution exists analysis

[binbagsss]

Homework Statement

Question:

For $\Lambda^3+\Lambda^2+A=0$ , show that for $A<0$ there is some real positive $\Lambda$ which solves this

Homework Equations

The Attempt at a Solution

[/B]

Attempt:

Write $-A=\hat{A}$, then $\hat{A} \in (0,\infty)$
and $\Lambda^3+\Lambda^2=\hat{A}=\Lambda^2(\Lambda+1)$

Since $\Lambda \in (-\infty, \infty)$ then $\Lambda^2 \in (0,\infty)$, and so since $\hat{A} \in (0,\infty)$ then $(\Lambda-1) \in (0,\infty) \implies \Lambda \in (-1,\infty)$

So this shift by $-1$ means that $\Lambda$ is not guaranteed to be positive?

Solution attached:

I understand the solution after the 'claim' $\Lambda \in (0,\infty)$.

However I don't understand where this comes from, this is part of what we are trying to show and we only have $\Lambda$ is real?

Many thanks in advance.

 

[Mark44]

Homework Statement

Question:

For $\Lambda^3+\Lambda^2+A=0$ , show that for $A<0$ there is some real positive $\Lambda$ which solves this

Homework Equations

The Attempt at a Solution

[/B]

Attempt:

Write $-A=\hat{A}$, then $\hat{A} \in (0,\infty)$
and $\Lambda^3+\Lambda^2=\hat{A}=\Lambda^2(\Lambda+1)$

Since $\Lambda \in (-\infty, \infty)$ then $\Lambda^2 \in (0,\infty)$, and so since $\hat{A} \in (0,\infty)$ then $(\Lambda-1) \in (0,\infty) \implies \Lambda \in (-1,\infty)$

So this shift by $-1$ means that $\Lambda$ is not guaranteed to be positive?

Solution attached:

View attachment 205800
I understand the solution after the 'claim' $\Lambda \in (0,\infty)$.

However I don't understand where this comes from, this is part of what we are trying to show and we only have $\Lambda$ is real?

Your post would be easier to read if you followed the lead of the material you posted, and used lower-case lambda ($\lambda$) instead of upper-case lambda ($\Lambda$). At first I misread $\Lambda$ as A. In TeX, the lower-case Greek letters have spellings that start with lower-case letters; i.e., \lambda versus \Lambda, and \theta versus \Theta.

Consider $f(\lambda) = \lambda^3 + \lambda^2 + A$, with A < 0.
$f'(\lambda) = 3\lambda^2 + 2\lambda$
On what intervals is f' positive? zero? negative? On these same intervals the original function will be increasing, horizontal, or decreasing, respectively. This should give you a good idea of where any zeroes of the graph of f are.

BTW, two of the intervals in the posted image use nonstandard notation-- $(0, \infty]$. I've always seen these written as $(0, \infty)$ when an interval extends infinitely far in either direction.

 

[LCKurtz]

I don't think it even requires calculus. Write it as $f(x) = x^3+x^2 -a$ where $a>0$. Then $f(0) = -a<0$ and $f(x) > 0$ for $x>a$.

 

[SqueeSpleen]

It doesn't requiere a lot of analysis, but you need to use intermediate value theorem (Bolzano's).

 

[WWGD]

Actually, this is true for any A, since Complex roots come in pairs, i.e., if z is a root, so is its conjugate z^ , so , by parity alone, there will be a Real root no matter the value of A.

 

[LCKurtz]

Actually, this is true for any A, since Complex roots come in pairs, i.e., if z is a root, so is its conjugate z^ , so , by parity alone, there will be a Real root no matter the value of A.

Yes, but will it be positive?

 

[WWGD]

Yes, but will it be positive?

Positive... you got me for not reading carefully; top-down processing, my bad. Good point. Maybe I should just delete my post?

 

[LCKurtz]

Nah. I wouldn't worry about it. It happens a lot to all of us.

 

[SqueeSpleen]

Also, some of the algebraics proofs of fundamental theorem of algebra uses the intermediate value theorem in one step (at least the ones that I know with Galois Theory). Then there's are complex analysis proofs that also uses analysis. I think that the less analysis one is the one that only uses the fundamental group of the circle, but intermediate value theorem is more a topological theorem than an analysis one. So I think that using fundamental theorem of algebra to avoid using bolzano's theorem is like a snake bitting it's own tail.

 

[LCKurtz]

Depending on the level of course in which the intermediate value theorem is used, it may very well be that the IVT is taken as "obvious" or an unproved given statement. A high school algebra student may "know" a "continuous" graph can't go from positive to negative without hitting the $x$ axis.

 

[SqueeSpleen]

Yes, you simply justify the afirmation talking about the graphic. But I was saying that it makes sense to put analysis in the tittle, due to that theorem. Of course, it's not mandatory.