An amazing way to determine the sign of a polynomial (1 Viewer)

[vafa]

As you know when you want to solve an iequality or when you need to determine the nature of stationary points, you need to work with numbers and test if the sign of the derivative changes before and after that point.

Recently i found a very interesting way and easy to do this. I studied this in a Book which is designed for talented mathematics student. the Author is Vafa Hedayati. He has the same first name as mine but obviously I am not him because he is a brilliant and I am so dumb.

in his method he says when you want to determine the sign of a polynomial, first u need to solve that polynomial to get the zeros of the polynomial then u will have 3 different cases but before that u need to arrange the roots from smallest to biggest.
if u have a linear polynomial the sign of the polynomial before the root is opposite the sign of the coefficient of x and after the root the sign of the polynomial is same as the sign of the coefficient of x.

if u have a polynomial of degree 2 then generally u may have 2 roots, in this condition the sign of the polynomial between two roots are opposite the sign of the coefficient of x^2 and in other places the sign is the same as the sign of coefficient of the x^2.

if u have cubic or quartic or more u need to break down it into quadric and linear, determine their sign and lastly multiply their sign to see what u get.

i have tested this method for many i can say about 100 polynomials and this method really works.
in my pespective this method is much better than method which we use. our method is really frustrating. u need to put values for x then calculate the value of the derivative while in this method u determine the sign of polynomial without testing and this method is much quicker.

 

[pLuvia]

The "Third Derivative Method" does the same thing, determines the nature of stationary point and points of inflexion

 

[joesmith1975]

hows the 3rd Derivative Method work??

 

[pLuvia]

Compliled by Buchanan take a look at this thread

 

[vafa]

Dear, Iruka here is my answer to your questions:
if u have a plynomial of degree 2 which has double root, obviously that is always greater than or equal to zero there is no exception.
if u have a plynomial of degree 5 , u need to get the zeros any way and this is done by factorisation. so u can do that kind of polynomial as well.
what about if u have a polynomial whose have double roots?
to determine the sign of the polynomial u look at the power of the expression which contains multiple root, if it is even u ignore that because u know that it is always zero or greater than zero but if the power is odd then u know that for x biger than multiple root that expression will be positive and for x less than multiple root that expresssion will be negative.
i agree that u may find this method very hard to understanding in the first step but the more u work with it, the more u will be love with it.
at least i have chosen this method for determining the sign of a polynomial.
i am not forcing everyone to do this way. all i am saying is we have this method and then that will be your choice to choose the best method.
i feel very ok with this new method and it always work. i had a look at buchanan method. i am sorry to say but there are some exceptions and this method does not work properly in all situations

 

[vafa]

Where Buchanan's method does not work:

His method will not work in cases where u have a function and in this function u have only x to the power of an odd number and also u have a constant.
e.g. f(x)=3x^5+1
if u look at my example first derivative will have x with a power which is an even number.
if u get the second derivative, the power of x will be an odd number
if u get the third derivative , the power of x will be an even number. now if u put xs which are negative and positive, the value which u will get will be always positive. so there is no point of infelexion...?
of course there is point of inflexion because if u look at second derivative, the power of x is an odd number and if u test number(i hate this frustrating method), the sign of the second derivative changes while by buchanan's method u would say there is no horizontal point of infelexion.
I need to mention that as buchanan has said the value which makes the second derivative is for example x if this x does not make the third derivtive zero, then that is a point of inflexion while if u look at my example x=0 makes second and third derivative zero so by buchahan's method there is no point of infelexion while in reality (0,1) is a horizontal point of inflexion.


remember something and this is Vafa Taher's method is not only for determining turning points or points of inflexion; u can use it for determining the sign of any polynomial. there is no exception for his method that is why i called him a brilliant because he created a method and fortunately there is no exception. so Well Done Mr.Vafa Taheri...!


can i just add that i do not want to unrespect to anybody or anybody's work. I just wanted to say that Buchanan's method does not work all the time. and if i have been rude, then i apologize for that.

 

[vafa]

if there is anybody who is interested to get Taheri's method, I can put in PDF and then make it avaliable for u guys to have a look at that but let me know because before doing this i need to get permission from him.

 

[Templar]

You still have not addressed the issue of resolving quintics and higher order polynomials into quadratic or linear factors where the roots might not be solvable (proven using Galois groups).

Also, you referred to the person as Vafa Hedayati at the start of the thread, and Vafa Taheri at the end. Neither yields any result with a Google search, which is highly suggestive that this person is entirely fictional, or at least his so called book for talented students does not exist. This is supported by the fact that the ideas presented are so elementary that anyone can formulate them.

And as a common courtesy, do not type in all bold, it is very straining on the eyes.

 

[vafa]

1) i never have said that his book is called maths for talented students
2) his name is Vafa Taheri Hedayati.
3) U can not say he is fictional by just looking at google search, search for me u can not find me there; so i am fictional...?
4) if u have a polynomial of degree 6, in order to know the sign of the polynomial u need to know its roots anyway. i think i just said that. and obviously in HSc they wont ask u a question where u can not find the roots.
5) remember that i have no reason to lie. if i wanted to lie, i would say this is done by me and by doing that i would be so proud of myself.
6) i am so sorry for u because u said that this is elementary. well if this is elementary, then whydid not u say that. if u did, your name would be recorded as a mathematician...!
7) this method is obviously not elementary. evidently he has spent a lot of time to test all possible situations to test if his method is ok.
8) if u think this method is very elemeneray and u r ina higher level of maths then plz make a formula. i will love to use your formula...!
9) always make sure that u r certain about something and then say your words because otherwise u will be so ashamed...!

 

[pLuvia]

But Templar is just implying since he made a book for talented students, his name should have at least be noted somewhere on the internet, either advertising his book etc.

 

[Riviet]

The method does sound intuitive, unfortunately, you can't really write this down on paper since it's not a theorem, just a shortcut/trick for determining the sign of polynomials. I guess you could use it as a quick check up, and use the normal methods that we are expected to use in the HSC.

 

[vafa]

well as i said this is yoyr choice to select your preferred approach.

however i am going to use this method for my hsc exam, i di it for my half yearly in maths extension 2 and maths extension 1 and my teacher said that is ok as long as u got the right answer.
i am using this method because i do not want to know maths like a primary school maths student.

 

[buchanan]

Compliled by Buchanan take a look at this thread

Thanks pLuvia.

Where Buchanan's method does not work:

His method will not work in cases where u have a function and in this function u have only x to the power of an odd number and also u have a constant.
e.g. f(x)=3x^5+1

Evidently vafa didn't read my thread very closely.

Here's my solution using my method.

f(x)=3x⁵+1

f'(0)=f''(0)=f'''(0)=f''''(0)=0 but f'''''(0)=3(5!)=360≠0.

∴ (0,1) is a horizontal point of inflection.

 

[YBK]

lol, I found out that method while doing problems myself a couple of years ago
is this what you mean:

2 - x^2

is concave down because x is negative?



That was pretty obvious

 

[vafa]

is that your derivative or it is just your f(x)?

 

[dom001]

Are you just talking about sign diagrams?
This new 'advanced' method is basically just common sense.
If you hvae all the real zeroes of the derivative and know whether they are single or repeated (either odd or even), then constructing the diagram is basically logic. If the derivative = 0, there is a horizontal tangeant (either min/max or horizontal inflection).
Now, obviously, if it is an even repeated root, it will be a min/max.
Ie, the graph of the derivative will touch the axis. Hence, the sign (pos/neg) around this point wont change.
If it is either an odd repeated root or a single root, the graph will cut the axis. Ie, the sign of the graph will change from pos to neg or neg to pos.
Theres nothing special or advanced about it. But if it saves you a bit of time, then use it.

 

[YBK]

is that your derivative or it is just your f(x)?

ummm.. i meant y=2-x^2

which is f(x)

 

[.ben]

what if some roots are complex

 

[acmilan]

'Brilliant' is subjective i guess...