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Polynomial - Newton's Method (3 Viewers)

Lukybear

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Using Newton's Method, under what conditions will a better estimate NOT be reached?
 

Trebla

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If the initial estimate is at a stationary point or if there is a turning point between the initial estimate and the actual root...
 

cutemouse

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Or if at P(z1) and P''(z1) have opposite signs, where x=z1 is the first approximation.
 

cutemouse

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jm01's case just means there is no root to find in the vicinity of your first approximation.
That is wrong.

My analysis involves the concavity and the sign of the curve (ie. below or above the x axis).

Taking the case that the curve that we're dealing with is concave up, the first approximation can still be near the actual root but the point on the curve could lie under the x axis. This would mean that Newton's method would not had been effective in giving a better approximation of the root.
 

cutemouse

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As you've not mentioned tangents anywhere in your reasoning, I can only assume you don't understand how Newton's method works.
And I guess because you didn't mention the word "mathematics" in YOUR argument I guess then you don't have a clue about what you're saying :rolleyes:

The signs of P(z1) and P"(z1) need to be the same for a good 1st approximation using Newton's Method for finding roots of a polynomial.

Suppose we have a parabola, say .

The tangent drawn from say x=3 will cut the x axis further away from the root than the tangent drawn from say x=4. Is this the "reasoning" that you desire? :rolleyes:
 
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Iruka

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And I guess because you didn't mention the word "mathematics" in YOUR argument I guess then you don't have a clue about what you're saying :rolleyes:

The signs of P(z1) and P"(z1) need to be the same for a good 1st approximation using Newton's Method for finding roots of a polynomial.

Suppose we have a parabola, say .

The tangent drawn from say x=3 will cut the x axis further away from the root than the tangent drawn from say x=4. Is this the "reasoning" that you desire? :rolleyes:
Why don't you get a decent CAS and actually have a look at your *example*?
 

Trebla

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And I guess because you didn't mention the word "mathematics" in YOUR argument I guess then you don't have a clue about what you're saying :rolleyes:

The signs of P(z1) and P"(z1) need to be the same for a good 1st approximation using Newton's Method for finding roots of a polynomial.

Suppose we have a parabola, say .

The tangent drawn from say x=3 will cut the x axis further away from the root than the tangent drawn from say x=4. Is this the "reasoning" that you desire? :rolleyes:
The example you provided is flawed.

(i)




(ii)



Which one do you think is a better approximation? :p
 
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Lukybear

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What about say y=x^(1/3). Isnt there some restriction on how the gradient of the tangent vs the gradient of the graph? I.e. at some pts on the graph, near the root, im sure it wont acquire better estimate.

Note: its due to tangent reaching a pt on the x axis, that is further away from the root then original estimation, due to graph sloping up quickly.
 
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cutemouse

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The example you provided is flawed.
Maybe I messed something up because I did it mentally.

But consider something like say

Why would x=2.1 have been a worse choice as a first approximation using Newton's Method of estimating polynomial roots over x=2.3?
 

Iruka

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Maybe I messed something up because I did it mentally.

But consider something like say

Why would x=2.1 have been a worse choice as a first approximation using Newton's Method of estimating polynomial roots over x=2.3?
The root of that polynomial is approximately 2.214014591. The reason that 2.3 gives a closer estimate than 2.1 is that it lies closer to the value of the root. It is not because it lies above the x-axis.
 

Iruka

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Yes, Newton's method does not guarantee convergence to the closest root in all cases. But the idea is that if the initial guess is *close enough* to the root, and the root it not at a stationary point, the method will converge. (Unfortunately, it can be very hard to figure out how close close enough is.)
 

cutemouse

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The original question said "Using Newton's Method, under what conditions will a better estimate NOT be reached?"

In your example, whether I start from 2.1 or 2.3, a better estimate is reached with the next step.
If you start from the 1st approx x=2.1 a better estimate is not reached.

The root of that polynomial is approximately 2.214014591. The reason that 2.3 gives a closer estimate than 2.1 is that it lies closer to the value of the root. It is not because it lies above the x-axis.
lol. You just contradicted yourself in the same sentence.

You said that the actual root was x=2.21... And you said that x=2.3 is closer to the root than x=2.1? They're pretty much about the same (I think)...

And plus the closeness to the root wouldn't matter (in a significant manner) anyway. It's all to do with the signs of P(z1) and P"(z1) usually needing to be the same for a good 1st approximation.
 
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Trebla

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You said that the actual root was x=2.21... And you said that x=2.3 is closer to the root than x=2.1? They're pretty much about the same (I think)...
Um...the distance between 2.10 to 2.21 is 0.11 whereas the distance between 2.21 and 2.30 is 0.09. I'd hardly call that a self-contradiction. We're talking on a very small scale here since both estimates are close to the actual root and clearly 2.3 is closer to the real root than 2.1 which means that after applying Newton's method with the initial approximation of 2.3, this gives a better estimate of the root.

Also, the parabola example you provided at first already disproved your statement, therefore that statement cannot be asserted as true across all general functions even if you provide specific functions that actually do satisfy what you said.
 

cutemouse

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I haven't looked into that parabola example yet. I'll do it tomorrow or something.

But it's not like it came to me in a dream or anything. It's published in one of the books of the perhaps one of the most prolific HSC Maths authors, ie. Jim Coroneos' 3U book. So it can't really be wrong.
 

study-freak

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I haven't looked into that parabola example yet. I'll do it tomorrow or something.

But it's not like it came to me in a dream or anything. It's published in one of the books of the perhaps one of the most prolific HSC Maths authors, ie. Jim Coroneos' 3U book. So it can't really be wrong.
No one is perfect.

That's like saying Conquering Chemistry/Jacaranda Chemistry can't have anything wrong in it. They can. (Not to mention that I've found some errors in Jacaranda Chemistry during my HSC)
 
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cutemouse

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No one is perfect.

That's like saying Conquering Chemistry/Jacaranda Chemistry can't have anything wrong in it. It can. (Not to mention that I've found some errors in Jacaranda Chemistry during my HSC)
Yeah I know what you mean but I believe that Coroneos had his booked checked by alot of people before publication. Plus most content in the books have been published since something like 1961... that's a long time compared to many other books... So the chance of any error is pretty slim.
 

study-freak

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Yeah I know what you mean but I believe that Coroneos had his booked checked by alot of people before publication. Plus most content in the books have been published since something like 1961... that's a long time compared to many other books... So the chance of any error is pretty slim.
Well, the first edition of Excel Physics was published quite a long time ago. Not as long ago as 1961 but still, since it has errors in it, there's Coroneos might have some too.

Yes, slim chance but it still can have some.
 

Iruka

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I have just checked Coroneous' Year 12 3 unit book, and he does rabbit on a bit about the signs of f(x_0) and f''(x_0), (it is on pp267-268 of the edition that I happen to have).

I would like to draw your attention to the following quote (which is italicized in the original):

"However, in all cases, if the first approximation is near to the true value of the root, then the second approximation will be nearer, and so on."

So I am afraid that even the esteemed Coroneous is agreeing with what everyone else here has been saying - getting a first approximation that is close to the root is the most important step in implementing the algorithm. More important than worrying about the signs of f(x_0) and f''(x_0).
 

shaon0

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Oscillating functions can cause problems when using Newton's method.
 

cutemouse

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Coroneous'
=\ spell the guy's name correctly: Coroneos.

So I am afraid that even the esteemed Coroneous is agreeing with what everyone else here has been saying
I never really encountered a question that involved the closeness of a root. I shall have a look later but nevertheless the signs of P(z1) and P"(z1) do matter, which is what I said initially.
 

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