Quadratic equation and its roots

In summary, the calculator always converges to the same solution when you repeatedly enter 1+1/Ans into it.
  • #1
kshitij
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Homework Statement
What are the roots of the equation x=1+1/x?
Relevant Equations
x=(-b±√(b²-4ac))/(2a),
when ax^2+bx+c=0
On simplifying the given equation we get, x^2-x-1=0 and using the quadratic formula we get x=(1+√5)/2 and x=(1-√5)/2
Now, as the formula suggests, there are two possible values for x which satisfies the given equation.

But now, if we follow a process in any general calculator by entering
1+1/Ans
into the calculator repeatedly, we always reach this particular number as shown,
1613129272919.png

Now obviously we cannot do this process for when the value of Ans is set to 0.
But other than that, no matter what the value of answer we choose,
for all real numbers if we repeatedly do this process on any calculator,
we always reach the above mentioned number.

This still somehow makes sense to me because if we are repeating our input into the equation 1+1/Ans
what we are actually doing looks somewhat like this,
Ans=1+1/Ans
here Ans is variable

So we eventually did end up with our starting equation whose roots were,
(1+√5)/2 and (1-√5)/2
And the number which the calculator shows is also equal to (1+√5)/2 But my question is why do we end up always at (1+√5)/2 and and never on (1-√5)/2?
x=(1-√5)/2 also satisfies the original equation so why does the calculator never show us the other root of this equation?

Now this got me thinking and I tried to manipulate the original equation to something like this (don't ask me why I did this cause even I don't know that, I was just playing around with random inputs like these when the result for this input surprised me),
x=1/(x-1)
This equation is exactly same as our original equation with the only difference being that in the original equation we cannot have the value of x as 0 and in the modified version, x should not be equal to 1. So now let's do another process (repeatedly) in the calculator,
1/(Ans-1)

again by the same logic we can see that the process which we are doing, actually looks like the equation,
Ans=1/(Ans-1)
Only difference this time is that now we cannot set the value of answer as 1 or 2.
Other than that, for all real values of Ans, this time we reach this particular number,
1613131014410.png

And this number is exactly equal to the other value of x in our original equation which is (1-√5)/2

So that is my question, why does the calculator have a favorite root for a particular function when more than one values satisfy the equation?
If both the roots are equivalent, then half of the times we should have reached (1+√5)/2 and other half of the times (1-√5)/2 randomly.

Also this isn't some usual homework question, I didn't knew whom to ask this question so I posted it here. I hope it doesn't break any rules.
 
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  • #2
Hi,

kshitij said:
But now, if we follow a process in any general calculator by entering
1+1/Ans
into the calculator repeatedly, we always reach this particular number as shown,

kshitij said:
But other than that, no matter what the value of answer we choose,
for all real numbers if we repeatedly do this process on any calculator,
we always reach the above mentioned number.
You want to be careful with generalizations.

Basically what you are doing is you solve x = x + 1/x by a well-known procedure called successive substitution: you solve x = f(x) by starting at some x0 and calculating f(x0). This result is your next guess, x1 and you calculate x2=f(x1).
Etc. until your x doesn't change any more.

Example:
x0 = 3
x1 = 1.333
x2 = 1.747
x3 = 1.571 etc converging to (1+√5)/2

1613135729904.png


and sure enough, x0 = -3 brings you to the same point.
BUT do you see which starting values take you to the alternative solution ?

Hint: It's only a small range. But your generalization 'no matter what the value of answer we choose' is definitely incorrect !

Advanced question: can you also see what kind of functions/starting value will not converge with this method ?

kshitij said:
Also this isn't some usual homework question, I didn't knew whom to ask this question so I posted it here. I hope it doesn't break any rules.
It is a very good and clear question!
 
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  • #3
Here you see the situation for x = 1/(x-1)

1613137001871.png

Again: can you find which starting values take you to the alternative solution ?

kshitij said:
If both the roots are equivalent, then half of the times we should have reached (1+√5)/2 and other half of the times (1-√5)/2 randomly.
Conclusion: both roots are almost never equivalent (never say never :wink: unless you're sure)

##\ ##
 
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  • #4
BvU said:
can you find which starting values take you to the alternative solution ?
It looks like,
for the first graph only the input (1-√5)/2 will get us to the alternate solution,
and similarly for the second graph only (1+√5)/2 gives us exactly the intersection between the lines y=x and y=1/(x-1),
anything other than that seems to slowly-slowly move away from (1+√5)/2.
That's when I tried to manually trace straight lines on the given graphs. So, I'm not sure but when I was trying this process on the calculator I did end up on the alternate solution every-time I started with the alternate solution.

edit: I will surely try this on GeoGebra as well.
 
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  • #5
BvU said:
can you also see what kind of functions/starting value will not converge with this method ?
I can obviously see that when you get close to an asymptote (x=0 in the first graph and x=1 in the second), it will not converge. Now what are the possible inputs that will reach to an asymptote, that I'm not sure about.

Obviously starting on an asymptote surely doesn't converge, but there are other values too (e.g- x=2,1.5 in the second graph) that ends up on an asymptote, But I don't how to find them. I'll think about it.

edit: I think I found a pattern to reach an asymptote,
lets talk about the second equation (because this method doesn't work for the first equation)
y=1/(x-1)
we just solve for the values of x where the equation gives us y=1
i.e., 1/(x-1)=1
which gives x=2,

Now again if we find some input which will give 2 as output, the function will not converge as 2 will again give us back 1 as output.
so,
1/(x-1)=2
which gives x=3/2

Similarly, we can find lots of values namely,
x=1,2,3/2,5/3,8/5,13/8...

Also I spotted something pretty cool,
these values of x's are the ratios of consecutive Fibonacci numbers!
Which is cool because I've heard that the ratio of consecutive Fibonacci numbers approaches the golden ratio!
the golden ratio is basically our solution (1+√5)/2

This is crazy because if we start with the golden ratio then it converges again!

Here's a gif where I plotted few of these values of x
Webp.net-gifmaker.gif
 
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  • #6
BvU said:
Here you see the situation for x = 1/(x-1)

View attachment 277816
Again: can you find which starting values take you to the alternative solution ?

Conclusion: both roots are almost never equivalent (never say never :wink: unless you're sure)

##\ ##
I almost forgot to thank you for your wonderful response! Those graphs do help a lot to understand the problem.
 
  • #7
kshitij said:
only the input (1-√5)/2 will get us to the alternate solution
You are absolutely right !

Bad on my part (shouldn't do this off the top of my head).
Has to do with the slope of f(x) -- more googling (or diving into the books :smile:) required .
E.g section 7.5http://math.jbpub.com/numericalmethods/pdf/Linz07.pdf
 
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  • #8
BvU said:
You are absolutely right !

Bad on my part (shouldn't do this off the top of my head).
Has to do with the slope of f(x) -- more googling (or diving into the books :smile:) required .
E.g section 7.5http://math.jbpub.com/numericalmethods/pdf/Linz07.pdf
I think this video is relevant here
 
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  • #9
This iterative method for solving x = f(x) will converge if there is a region around the root in which |f'(x)| < 1.
 
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  • #10
Chestermiller said:
This iterative method for solving x = f(x) will converge if there is a region around the root in which |f'(x)| < 1.
That is very interesting, in fact this video which I shared earlier also talks about how this has something to do with slopes.

But I can't quite get why is it so?

It somewhat feels like the concept of stable and unstable equilibrium in physics, where the root at which this function converges is at stable equilibrium and the other is unstable. This is just how I see it, but this doesn't explain why a region around the root in which |f'(x)| < 1 is the stable equilibrium position?

Also I have some other doubts as well.

I was plotting the same function by rearranging its terms, on GeoGebra, and pretty much everyone of them followed your rule

1613218979685.png
1613219027411.png

these above two were the original ones which I mentioned in the post,

And these two below were two similar cases which followed your rule,
1613219189444.png
1613219256609.png


I also found a few interesting cases as well like these below,
1613219364496.png
1613219418092.png

here we can see that,
|f'(x)| > 1 in the regions around both the roots,
so in these two cases our method should not work and it should not converge.

But there were some values for which our function didn't blow up to infinity, but instead it was dancing around two particular values.
Why is that happening?

Also my other question is,
can we have a function, for which |f'(x)| < 1 at regions around both the roots?
What will happen in this case? Will our function converge to both roots randomly?

Now I tried to guess one such function, but the one I found had some defects, it involved modulus and square roots,
##|y|= √ (x+1)##

So, for this I tried my process with two separate functions,
##y= √ (x+1)## and ##y= - √ (x+1)##
1613220144148.png


Now, this works for the one with positive sign but not for the one with negative sign, is this because of the square root which restricts our domain to only non-negative integers?

Anyway to sum it up,
1.) Why does our function converge if there is a region around the root in which |f'(x)| < 1?
2.) For functions in which |f'(x)| > 1 around both the roots, why are there some values for which f(x) ended up fluctuating between 2 particular values?
3.) Can we have a function where the regions around both the roots have |f'(x)| < 1? Will this function converge randomly to either of our roots in our process?
 
  • #11
kshitij said:
But I can't quite get why is it so?
Did the exposé on p 190 in the link I gave earlier not explain that clearly ?
 
  • #12
BvU said:
Did the exposé on p 190 in the link I gave earlier not explain that clearly ?
I actually didn't read that because when I started with section 7.5, a lot of stuff went over my head, I thought that I might have to read the whole thing from beginning or maybe my knowledge isn't enough to understand all of that, so I just saved that document for now.

I can still give it another try though.
 
  • #13
Start with a linear recurrence: By induction, you can show that the solution of [tex]a_{n+1} = \lambda a_n[/tex] is [tex]a_n = a_0\lambda^n[/tex]. Now [itex]|a_n| = |a_0||\lambda|^n[/itex] is strictly decreasing if and only if [itex]|\lambda|^{n+1} < |\lambda|^n[/itex], ie. [itex]|\lambda| < 1[/itex].

Now consider the non-linear recurrence [tex]
x_{n+1} = f(x_n)[/tex]
Near a fixed point [itex]x_0[/itex] we can set [itex]x_n = x_0 + \epsilon_n[/itex] and expand [itex]f(x_0 + \epsilon_n)[/itex] in as a taylor series: [tex]
x_0 + \epsilon_{n+1} = f(x_0) + \epsilon_nf'(x_0) + \cdots
[/tex] Since [itex]x_0 = f(x_0)[/itex] this leaves [itex]\epsilon_{n+1} \approx f'(0)\epsilon_n[/itex]. The linear theory tells us that [itex]|\epsilon_n|[/itex] will increase if [itex]|f'(x_0)| > 1[/itex] and decrease if [itex]|f'(x_0)| < 1[/itex]. If [itex]f'(x_0) = 1[/itex] then this test is inconclusive.
 
  • #14
pasmith said:
Start with a linear recurrence: By induction, you can show that the solution of [tex]a_{n+1} = \lambda a_n[/tex] is [tex]a_n = a_0\lambda^n[/tex]. Now [itex]|a_n| = |a_0||\lambda|^n[/itex] is strictly decreasing if and only if [itex]|\lambda|^{n+1} < |\lambda|^n[/itex], ie. [itex]|\lambda| < 1[/itex].

Now consider the non-linear recurrence [tex]
x_{n+1} = f(x_n)[/tex]
Near a fixed point [itex]x_0[/itex] we can set [itex]x_n = x_0 + \epsilon_n[/itex] and expand [itex]f(x_0 + \epsilon_n)[/itex] in as a taylor series: [tex]
x_0 + \epsilon_{n+1} = f(x_0) + \epsilon_nf'(x_0) + \cdots
[/tex] Since [itex]x_0 = f(x_0)[/itex] this leaves [itex]\epsilon_{n+1} \approx f'(0)\epsilon_n[/itex]. The linear theory tells us that [itex]|\epsilon_n|[/itex] will increase if [itex]|f'(x_0)| > 1[/itex] and decrease if [itex]|f'(x_0)| < 1[/itex]. If [itex]f'(x_0) = 1[/itex] then this test is inconclusive.
I'm so sorry but this is beyond me. I thought that there must be some geometric way to see this through graphs, like the one in the first two responses by @BvU. I appreciate your efforts but I think this question is not for me (atleast not for now), But I do hope to understand this someday.

Thank you for your response.
 
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  • #15
Suppose that the exact root is ##x_E##. The recursion relationship being used is $$x^{n+1}=f(x^n)$$If we expand the right hand side in a Taylor series about ##x_E## and retain the linear terms, we obtain: $$x^{n+1}=f(x_E+x^n-x_E)=f(x_E)+f'(x_E)(x^n-x_E)$$From this it follows that $$(x^{n+1}-x_E)=f'(x_E)(x^n-x_E)$$and$$\frac{|x^{n+1}-x_E|}{|x^n-x_E|}=|f'(x_E)|$$So the magnitude of the deviation from the exact answer is getting smaler from one iteration to the next if the absolute value of the derivative of the function is less than unity at the root.
 
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Related to Quadratic equation and its roots

1. What is a quadratic equation?

A quadratic equation is a polynomial equation of the second degree, meaning it contains a variable raised to the power of 2. It can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable.

2. How many solutions can a quadratic equation have?

A quadratic equation can have two solutions, one solution, or no real solutions. The number of solutions depends on the value of the discriminant, which is b^2 - 4ac. If the discriminant is positive, there will be two real solutions. If it is zero, there will be one real solution. And if it is negative, there will be no real solutions.

3. What is the quadratic formula?

The quadratic formula is a formula used to solve any quadratic equation. It is written as x = (-b ± √(b^2 - 4ac)) / 2a. The ± symbol means that there are two solutions, one with a plus sign and one with a minus sign.

4. How do you find the roots of a quadratic equation?

The roots of a quadratic equation are the values of x that make the equation equal to 0. To find the roots, you can use the quadratic formula or factor the equation into two binomials. If the equation cannot be factored, the quadratic formula is the best method to find the roots.

5. What is the relationship between the roots of a quadratic equation and its graph?

The roots of a quadratic equation are also known as the x-intercepts or solutions of the equation. On a graph, the x-intercepts are the points where the graph crosses the x-axis. The number of roots corresponds to the number of times the graph intersects the x-axis. If there are two roots, the graph will intersect the x-axis twice. If there is one root, the graph will intersect the x-axis once. And if there are no real roots, the graph will not intersect the x-axis at all.

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