Degrees of terms in a polynomial

[Dillinger4]

Can anyone explain why the degree of 9 in the polynomial 4x³ - 5x + 9 is 0? The explanation i get from an alternate source is this. 6 = 6x⁰. But then wouldnt 6¹ = 6 = 6x⁰? So why would 9 have a degree of 0 and not 1?. Thanks.

[sql_lall]

The degree of the "9" is 0

The degree of the "4" is 3, because it is 4* x^ 3
it is also 4^1 * x^3
BUT, the degree is what power the unknown term (x) is raised to.
Because 6=6*x^0, the degree is 0.

I.e. 6 is the CONSTANT term (=degree 0)

[Dillinger4]

Posted by sql_lall

BUT, the degree is what power the unknown term (x) is raised to.

I think i need further clarification.

In 4x3 - 5x + 9 the term -5x has a degree of 1 because -5x = -5 * x¹ Now you have stated that the degree is the power that the unknown term is raised to. But why are we suggesting that 9 = 9 * x⁰? Even though that is true why are we reforming the expression? Thanks for the help.

[Guv]

For many purposes, it is handy to be able to use expressions like the following

Polynomial(x) = Sum(A_j * x^j) for j = 0 to n

The above is a very compact notation for the following.

A₀ * x⁰ + A₁ * x¹ + A₂ * x² . . . + A_n * xⁿ, where A₀, A₁, A₂ et cetera are coefficients.

The compact notation requires us to think of the constant term as having zero as the exponent for x, and to think of the next term as having one as the exponent.

Does the above help any?

[Dillinger4]

Seems like to go further i had to go back. I rethought what a monomial is(a constant or a product of constants and one or more variables with whole number exponents). Now since a polynomial is the sum of one of more monomials the 9 in 4x³ - 5x + 9 should be in monomial format. 9x⁰. Which would have a degree of 0. Would this be correct?

[Guv]

Dilenger4: Yes, I think you are correct.

I do not remember the term monomial, although it must be a valid word.

It is not common practice to use zero or one as exponents. In some contexts, we should imagine that an exponent of zero or one is there.

[sql_lall]

In all honestly, being absolutely by-the-book, i guess you should always put 'x⁰' after the constant term. However, as this is =1, people just got lazy and left it off.
Interestingly though is that if x=0, you get 0⁰=1, which is debated over, and often determined to be undefined, not =1

[Guv]

0⁰ = 1 is correct and necessary for consistency. I cannot imagine who would define it otherwise.

[nishantp]

Originally posted by Guv
0⁰ = 1 is correct and necessary for consistency. I cannot imagine who would define it otherwise.

The source of that particular debate seems to be that saying 0[sup]0 is like getting something from nothing.

[TheAlchemist]

i believe it is about requireing that particular definition to be made to preserve the consistency of the whole idea of x⁰
just like 0! = 1 to preserve the consistency of the combination and permutation formulas
i think it also has something to do with proof by induction

[Dillinger4]

That is one thing about mathematics that i find quite annoying. In order to maintain consistency it is often necessary to accept somthing that is perceived to be incorrect. ie..... 0⁰ = 1

How can 0¹ = 0 or 0² = 0 but 0⁰ = 1?

[Dillinger4]

Posted by Guv

0⁰ = 1 is correct and necessary for consistency. I cannot imagine who would define it otherwise.

Im not so sure about this. . I tried four calculators and looked in 3 books and all state that 0⁰ is undefined.

[Guv]

Dilenger4:You are using second rate calculators and books that make erroneous statements.

I tried four calculators and looked in 3 books and all state that 0⁰ is undefined.

That is one thing about mathematics that i find quite annoying. In order to maintain consistency it is often necessary to accept somthing that is perceived to be incorrect. ie..... 0⁰ = 1

0⁰ = 1 when I calculate it using my HP 48GX, and I would be surprised if the high end TI calculators did not agree.

I hope that no text book being used in a school or college claims that 0⁰ is anything but one.

Nonsense and opinion have no place in mathematics. Opinion and nonsense are given more credibility than they deserve in pseudo-sciences like Economic Theory, Political Science, Psychology, and some others. In those disciplines, this is understandable due to the nebulous nature of those subjects. Still, the authorities in those subjects should be more willing to admit to lack of knowledge instead of presenting unprovable concepts as valid science.

Mathematicians have damn good reasons for various counter intuitive statements. Let us analyze 0⁰ by considering x⁰, where x is real. We should also consider z⁰ where z=x+iy, a complex number.

What values does x⁰ have as x approaches zero from either the positive or negative direction? Let us build a table of some of those values.

• 100⁰ = 1
• 10⁰ = 1
• 1⁰ = 1
• .5⁰ = 1
• .01⁰ = 1
• .0000001⁰ = 1
• .000000000000000000001⁰ = 1
• Similarly for negative values of x.

Do you really expect x⁰ to be discontinuous at x = 0 ? Can you think of any value other than one which would be reasonable?

A complex function can be plotted using ordinary 3D Cartesian coordinates. If you view z=x+iy as a point in the XY-Plane, Function(z) can be plotted using the vertical coordinate.

The plot of z⁰ is a plane one unit above the XY-Plane. Along any path toward z=0, all the values of z⁰ are on that plane. Do you expect the function to jump off the plane when z=0? Do you some other value at z = 0+i0 ?

There is more than consistency and opinion working here. There is logic and common sense.

[TheAlchemist]

i think Guv has really clarified one of the statements i made earlier:

i think it also has something to do with proof by induction

it is by observation, exactly the way guv put it, that this has been defined.

[Dillinger4]

I found a suitable explanation that makes sense to me.

(a*a*a)/(a*a*a) = a^(3-3) = a^0

a^3/a^3 = 1, and so a^0 = 1


Thanks for the help guys.

[sql_lall]

Guv:
"Do you really expect x⁰ to be discontinuous at x = 0 ? Can you think of any value other than one which would be reasonable? "

Think of x/x - 1. Now, for every real value of x EXCEPT x=0, this equates to 0. However, it IS discontinuous at x=0.

Dilenger4
"(a^3)/(a^3) = a^(3-3) = a^0 = 1"
What about if a = 0. you get 0^3/0^3 = 0/0 = 1??

I think that similar to dividing, this is discontinuous when talking about zero.

A thought.
0^x=0
y⁰=1
So, when you combine the two, why does one take precendence over the other? Surely, either it is 'both' (like the infinite sum of 1-1+1-1+1-1+...) or it is undefined.

[Dillinger4]

"(a^3)/(a^3) = a^(3-3) = a^0 = 1"

sql_lall
What about if a = 0. you get 0^3/0^3 = 0/0 = 1??

Opps sorry. I forgot to account for a having a value of 0.

But then that would equate to an undefined operation. Which i had previously stated.

What is 0 to the power 0?

Answer:
0^0 is usually considered as being indeterminate since there is no consistent way to define it. In other words, the 0⁰ cannot be defined to be consistent with the other properties of the real numbers and thus mathematicians choose not to define it.
0⁰ = (1 - 1)⁰, which you then may expand by the binomial theorem to obtain a limit of 1. Or

0⁰ = ⁰(1 - 1) = 0¹/0¹ = 0/0, which is not defined. Or

If we graph y = x^x, the curve approaches y = 1 at x = 0 (from the positive side). And
0² = 0; 0¹ = 0; 0^(1/2) = 0; 0^(-1/2) = 1/0 which is infinity.

At what point does the result jump up from 0 to infinity?

The answer(?):

0⁰ = 1 from the application of the binomial theorem. Thus it equals 1 in certain contexts.

The function has a discontinuity at (0,0) and as a result it is indeterminate.

[Guv]

Functions which evaluate to 0 / 0 for some critical value of a variable are sometimes deliberately defined to a noncontinuous value at that critical value of the independent variable. This is usually pedantry used to indicate the nature of the definition of a function. For example: F(1) = 2, F(2) = 5000, F(X) = 0 for all other values fits the technical definition of a function. You can define almost anything to be a function. However, standard analytic functions are defined to be equal to their limit values for critical values of the independent variable. Any other definition leads to inconsistency.

Sql_Lall: You are making a very common error.

Think of x/x - 1. Now, for every real value of x EXCEPT x=0, this equates to 0. However, it IS discontinuous at x=0.

You do not analyze the behavior of a function at some critical value by evaluating the function for the critical value.

For example: sin(x) / x is 0 / 0 for x = zero, but every calculus text written in the past hundred years would assign limit[ sin(x) / x ] = 1 as x approaches zero. The function would be assigned the limit value and not considered undefined at x = 0

Similarly x / x - 1 = 0 at x = zero. Consider values as x approaches zero. Do not evaluate at zero.

0 / 0 is considered undefined because some functions ( like x / x - 1) approach one value and others [ sin(x) / x ] approach another value.


Dilenger4: Where are you getting the idea that 0⁰ is undefined?

0^0 is usually considered as being indeterminate since there is no consistent way to define it. In other words, the 0⁰ cannot be defined to be consistent with the other properties of the real numbers and thus mathematicians choose not to define it.

0⁰ = 1 is consistent, and no serious mathematician would consider it undefined. Your analysis is faulty.

The 4 calculators and 3 books you mentioned in a previous post are wrong, erroneous, errant, claptrap!

Read a previous post of mine which shows that limit of (x+iy)⁰ = 1 for any path in the complex plane. Similarly for real x, when y = 0

Exactly which four calculators did you test this on? I want to make sure that none of my friends buy those calculators.

What three books told you that 0⁰ was undefined?

I would really like to check your sources of misinformation. If those books are used as textbooks, the publishers should be notified that they are teaching nonsense, and perhaps they should be reported to some authority.

[sql_lall]

"Interestingly though is that if x=0, you get 0⁰=1, which is debated over, and often determined to be undefined, not =1"
lol

Anyway:
"Similarly x / x - 1 = 0 at x = zero. Consider values as x approaches zero. Do not evaluate at zero"

i.e. you are saying that as X gets close to 0, x/x-1 is 1.
However, we are not talking about values of X CLOSE to 0, we are talking about values when X is EXACTLY 0.
For instance, if i asked you what 0/0 is, the answer is not "1, because as X gets really small X/X = 1" It is in fact "undefined, as you are trying to divide by 0"

The whole thing about calculus texts saying that sin(x)/x = 1 for x=0, and leaving it at that, is because you don't actually use the x=0 case when doing calculus. You are always testing the slope for some minute non-zero x-step. If it was truly an x-step at zero, then there would be no slope at all. (A point has no slope)

So, similar to the way 0/0 = undefined, 0⁰=undefined

BTW, if u used the 'limit' reasoning to show 0⁰=1, then couldn't u use the same reasoning to show 0⁰=0 (considering the limit of 0^x as x->0 from above)??

[Guv]

Dilenger4: I am still waiting for you to identify the 4 calculators and the 3 books mentioned in a post by you.

Sql_Lall: The last part of your most recent post is a reasonable argument. It brings up an issue that I had not thought about while making my previous posts.

BTW, if u used the 'limit' reasoning to show 0⁰=1, then couldn't u use the same reasoning to show 0⁰=0 (considering the limit of 0^x as x->0 from above)??

Various other arguments posted by you and others have no merit. I would only be repeating previous posts if I addressed them.

The above quoted argument is cogent. Frankly, I do not remember enough to know exactly what is wrong with it. The bolded (by me) part of the quote suggests the problem.

0^x is obviously discontinuous at x = 0.

For positive non-zero values of x, the value of the function is zero. EG: 0² = 0

For negative non-zero values of x, the value is the reciprocal of zero, which is worse than 0/0

EG: 0^-2 = 1 / 0²

The discontinuity is a serious flaw in the function.

x⁰ has no such flaw. The only questionable value of x⁰ is at x = 0

Notice that for complex as well as real values, A^-z = 1 / A^z

Therefore 0^-z = 1 / 0^z

The above indicates that 0^z is a pathological function, with more than one critical value of z.

The function z⁰ has no critical value other than at z = 0 It is obviously continuous if the value at z=0 is defined to be one. No definition of 0^z at z=0 results in a continuous function.

A mathematician currently taking serious courses would probably come up with a better counter argument than what I have presented.

z⁰ is not in the same class as either 0/0 or 0^z It has a reasonable and consistent value at z

[Dillinger4]

This is the older issue from 1993 so i dont know if the statements about 0⁰ have been changed in the new(1997) eddition which barnes & noble does have in stock.

http://search.barnesandnoble.com/Oop...=2819517039018

I doubt this book is being used to teach algebra in school but...

http://search.barnesandnoble.com/boo...64142140&itm=1


Page 124 states:

a⁰ = 1
a = any number except zero
0⁰ undefined


The book below is what is currently being used to teach algebra at my college(Bergen Community College).

http://search.barnesandnoble.com/tex...10&TXT=Y&itm=4

Page 306 states:

a⁰ = 1 for all a != 0 Note: 0⁰ is undefined.

As for calculators: 0^0 on the Sharp Scientific Calculator EL-506L
Advanced Dial produces Error 2.

I did a search on Sharps site and i cant seem to find that exact model so im not sure if it's been discontinued.

http://search.sharpusa.com/search/?s...p-k=&x=55&y=14

Next is the Texas Instruments TI-83 which produces the same results.

[Guv]

Dilenger4: After doing a bit of research, I have concluded that you are correct.

Your argument relating to 0^x is valid. 0⁰ is undefined.

Sin(x) / x is not sufficient to establish 0 / 0 as defined. Neither is x⁰ sufficient to establish 0⁰ as defined.

Unlike some at this and other forums, when I know I am wrong, I admit it. When I do not know I am wrong, I will continue arguing as was the case here.

[Dillinger4]

At least we cleared that up. Thanks for all the help from you guys.