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#1 2015-04-24 23:18:27

niharika_kumar
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From: Numeraland
Registered: 2013-02-12
Posts: 1,062

quadratic polynomial

Let P(x) be a quadratic polynomial with real coefficients such that for all real x the relation 2(1+P(x))=P(x-1)+P(x+1) holds.
P(0)=8 and P(2)=32.
If the range of P(x) is [m,∞), then the value of m is?


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#2 2015-04-24 23:31:21

bob bundy
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Registered: 2010-06-20
Posts: 8,151

Re: quadratic polynomial

hi Niharika,

You could proceed like this:

Use the 'P' recurrence relationship to work out P(1).

Then call the quadratic ax^2 + bx + c, substitute in the known values when x = 0, 1 and 2 and solve for a, b, and c.

Once you know the quadratic you'll see it has a positive x^2 so it is 'U' shaped rather than the other way up.  So it has a minimum value.  The minimum for a quadratic is always at -b/(2a) and all you need is the 'y' coordinate at that minimum point.

Hope that helps,

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#3 2015-04-25 00:33:22

Olinguito
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Registered: 2014-08-12
Posts: 649

Re: quadratic polynomial

straightaway.


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#4 2015-05-01 15:09:12

niharika_kumar
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From: Numeraland
Registered: 2013-02-12
Posts: 1,062

Re: quadratic polynomial

Here is one more question on quadratic equations.

Assume that p is a real number.Find the possible values of p in order to have real solutions for

.




I shifted x^1/3 to RHS and cubed both sides and tried to simplify it to a 2 degree equation so that I could apply D≥0 for real roots, but was unsuccessful in making the equation quadratic.
Pls help.


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#5 2015-05-01 17:07:04

bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: quadratic polynomial

Hi;

I tried the substitution x = y^3 and that found the value of the discriminant for p but I am not sure about that idea.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#6 2015-05-01 20:10:06

bob bundy
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Registered: 2010-06-20
Posts: 8,151

Re: quadratic polynomial

hi Niharika

I started by having a look at a graph.  By using 'a' instead of 'p' you can use the slider to try varying 'a'.

You'll see they all look similar with the intersection moving right as 'a' gets bigger.

Here's the graph when a = 2.

kCmTlq0.gif

So then I did this:

where y = x^(1/3)

So

We need y  ≥ 0 so

That seems to agree with the graphical results.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#7 2015-05-02 03:54:25

niharika_kumar
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From: Numeraland
Registered: 2013-02-12
Posts: 1,062

Re: quadratic polynomial

Well the answer given is p≥ -1/4.


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#8 2015-05-02 04:37:48

bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: quadratic polynomial

That is the answer I got but I do not like it.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#9 2015-05-02 05:09:49

bob bundy
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Registered: 2010-06-20
Posts: 8,151

Re: quadratic polynomial

That appears to come from the discriminant of the y quadratic.  But you still have to find x values from those ys.

If p = -0.25 then y = -0.5 and so x = -0.7937.

In the original equation the LHS = 0.10969 which is not 1.  So that is not a solution.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#10 2015-05-02 23:06:06

Olinguito
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Registered: 2014-08-12
Posts: 649

Re: quadratic polynomial

bob bundy wrote:

If p = -0.25 then y = -0.5 and so x = -0.7937.


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#11 2015-05-02 23:18:22

Olinguito
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Registered: 2014-08-12
Posts: 649

Re: quadratic polynomial

niharika_kumar wrote:

.

Last edited by Olinguito (2015-05-04 00:09:46)


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#12 2015-05-05 00:08:47

bob bundy
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Registered: 2010-06-20
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Re: quadratic polynomial

hi

Many thanks Olinguito, for finding my error.  I had done cube root of y to get x, instead of cube y to get x.

I was still puzzled about why my graph didn't show any negative x values.  I experimented with the grapher for some time but couldn't get a reliable graph.  As negative values will have a real cube root there should be a graph for negative x.  It seems to be the way the grapher has been programmed.

So I used Excel (sorry folks shame ) to generate values, checked a few by long hand, and when satisfied it was working out ok I made a scatter graph of the points.  That's the best graph I think, for this data.

Here's a screen shot of two graphs with p = -0.24 and -0.26 and then the critical one with p = -0.25. 

sTjQKVg.gif

nHuZSTO.gif

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#13 2015-05-05 10:34:53

Olinguito
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Registered: 2014-08-12
Posts: 649

Re: quadratic polynomial

You are taking values of x and p independently of each other – but they are not independent of each other! x and p are related by

In other words your <cube root c> minus <cube root a> must be equal to 1 – but as you can see they are not. shame

Last edited by Olinguito (2015-05-05 14:08:21)


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#14 2015-05-05 10:57:52

Olinguito
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Registered: 2014-08-12
Posts: 649

Re: quadratic polynomial

PS:

so maybe this is the graph you want to plot: http://www.wolframalpha.com/input/?i=pl … a=%5E_Real

Last edited by Olinguito (2015-05-05 14:09:18)


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#15 2015-05-05 21:14:24

bob bundy
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Posts: 8,151

Re: quadratic polynomial

hi Olinguito

I was deliberately taking x and p as independent; looking at the graph for a particular p, and finding which x (if any) makes the expression come to 1.  I explored a whole family of graphs with varying p and could see that as p approached -0.25 from above the graph dropped lower.  At p = -0.24 the graph just crosses the line y = 1.  At p = -0.25 it just touches y = 1; and at p = -0.26 the curve fails to rise high enough.

I realise that isn't a proof, but I like to get a visual demonstration of a result.  Before your proof others were unsure if p = -0.25 was the correct result.  I did the graphs to show it is plausible.  I'm following the bobbym's mantra to try it with numbers .

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#16 2015-05-06 07:18:55

bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: quadratic polynomial

bob bundy wrote:

Before your proof others were unsure if p = -0.25 was the correct result.

I am sorry for my comment without an explanation but I was unsure of the answer I got. For one thing, I do not like the way the problem is worded.

Find the possible values of p in order to have real solutions for

I am bothered by the word "possible."

Synonyms for possible: conceivable, plausible, imaginable, thinkable, believable, likely, potential, probable, credible, tenable, odds-on

Holmes wrote:

I am accustomed to have mystery at one end of my cases, but to have it at both ends is too confusing.

I am wary of problems that have any ambiguity in the phrasing. Anyway, that led me to try to find the real roots of that equation when p = - 1 / 4 or ( - 1 / 4 ) <= p < 0. I was unable to find any real solutions. Wolfram can not either unless you make a choice between principal value of the root or real value of the root. One produces an answer, the other does not.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#17 2015-05-07 01:15:53

niharika_kumar
Member
From: Numeraland
Registered: 2013-02-12
Posts: 1,062

Re: quadratic polynomial

well I came up with the same solution yesterday as olinguito.
It was quietly easy.
Thanks for helping me out.


friendship is tan 90°.

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#18 2015-05-07 03:32:11

bob bundy
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Registered: 2010-06-20
Posts: 8,151

Re: quadratic polynomial

bobbym wrote:

( - 1 / 4 ) <= p < 0. I was unable to find any real solutions.

Does that mean that my graph for p = -0.24 in post 12 is wrong?

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#19 2015-05-07 09:29:58

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: quadratic polynomial

Hi;

I truly do not know, that is why I said I did not like my answer.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#20 2015-05-07 10:23:12

bob bundy
Administrator
Registered: 2010-06-20
Posts: 8,151

Re: quadratic polynomial

hi bobbym,

If p = -0.24 and x = -0.064

x + 3p + 1 = - 0.064 - 0.72 + 1 = 0.216 so cube root = 0.6

and cube root of x = - 0.4 so LH expression = 0.6 -- 0.4 = 1

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#21 2015-05-07 12:23:13

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: quadratic polynomial

Hi Bob;

It depends on how you interpret the cube root. Wolfram believes there are two interpretations. A principal cube root and a real valued one, just like with square roots. It is my understanding that the default one is the principal cube root. Using that with your p and q will return a complex number and not 0.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#22 2015-05-07 15:06:38

Olinguito
Member
Registered: 2014-08-12
Posts: 649

Re: quadratic polynomial

When you allow complex values, then every real number will have exactly three cube roots: one of them is real and the other two conjugate complex numbers. If you want only real solutions, you should tell Wolfram explicitly – otherwise it will give you its "principal value", which is not always the real root.

Last edited by Olinguito (2015-05-07 15:10:32)


Bassaricyon neblina

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#23 2015-05-07 15:18:10

Olinguito
Member
Registered: 2014-08-12
Posts: 649

Re: quadratic polynomial

Wolfram always returns a complex number as the principal value of the cube root of a negative real number. Presumably this is the complex number in the first quadrant of the Argand diagram.

Last edited by Olinguito (2015-05-07 15:27:54)


Bassaricyon neblina

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#24 2015-05-07 20:11:31

Olinguito
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Registered: 2014-08-12
Posts: 649

Re: quadratic polynomial


Bassaricyon neblina

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