Degree of a polynomial Puns

This was inspired by a question in Abstract Algebra by Dummit and Foote, where they asked me to find all monic irreducible polynomials in F_2[x] of degree 1, 2, and 4 then show that their product is x^(2^4)-x. Which made me wonder why? What is special about the numbers 1, 2, and 4? Then I realized that they are the divisors of 4. So I conjectured that x^(p^n)-x is the product of all monic irreducible polynomials of degree dividing n. I did a couple of cases to verify and they all seemed to work, so I figured there must be a simple proof that I can find online. After searching for a bit I couldn't find a COMPLETE proof of this statement, but it is definitely a known result. So over the weekend I thought hard about this and came up with a complete proof. Here is the link

https://docdro.id/gD4RMlV

I hope this will help others in the future understand this neat result in finite fields.

How would you find zeros for a polynomial with a degree of 4?

The equation: f(x)=-0.006x^4 + 0.140x^3 - 0.053x^2 + 1.79x

I know a zero is x=0, but how do I find the other ones, and are there any imaginary zeros? I am able to see an actual x-int on the graph, but how would I find that with this function? Thank you.

A while back I was reading a book called “The Outer Limits of Reason” by Noson Yanofsky. It deals with a number of math and science problems that are impossible to solve. In it, the author mentions that polynomials of the fifth degree or higher are impossible to solve using a general algebraic formula like the quadratic formula or the less used cubic formula. I found that interesting and went to google for more information. However, to my disappointment, most everything I have found online was written for math majors and assumed that they had a few semesters of group theory and Galois theory. Is there anyone who can explain this to me, a non math major. I’ve had a year of college level calculus, so I have some math knowledge, but I didn’t major in math, engineering, or physics so I don’t have a very deep background in higher mathematics. Is it possible to explain it to me in a way that I could understand?

Hey guys, I've decieded to study calculus by myself and I just started the chapter about limits. A doubt arose in an example that asked us to prove, using the precise definition of a limit, that:

[;\lim_{x\to\3} x*x = 9;]

and it proceeded as follows:

[;|x*x - 9|<\epsilon;] if [;|x - 3|<\delta;]

[;|x+3||x-3|<\epsilon;]

[;|x-3|<\frac{\epsilon}{|x+3|} = \delta;]

Then, it was stated that as we were concerned with values of x close to 3 "it's reasonable to assume that x is within a distance 1 from 3", and therefore |x+3| = 7 was a valid choice([;\delta=\frac{\epsilon}{7};]). For me, though, that choice of |x+3| seemed completely arbitrary, and I would like to know if there is a more precise way of determining a value for x, and therefore for delta, in cases like this.

A similar problem occurred in an exercise of the same kind that asked us to prove that:

[;\lim_{x\to-2} x*x - 1 = 3;]

I have a bunch of data points that I'm trying to get to generate a trendline polynomial equation. I don't know what degree the polynomial equation should be. Is there a way, to get the most accurate polynomial equation, I can have Google Sheets decide what degree the polynomial should be, instead of me choosing?

Thanks!

I am looking to find the Leave-One-Out error of each degree of polynomial 1-27. Can someone let me know what I am doing wrong and point me in the right direction please! I am very new to this stuff.

for i in range(28):
    poly = PolynomialFeatures(degree=i)
    X_poly = poly.fit_transform(X)
    lm = LinearRegression()
    lm.fit(X_poly, y)
    cv = cross_val_score(lm, X, y, scoring=mse, cv=n)
    print(cv.mean())

https://imgur.com/DjE7oVR

I've seen numerous examples of similar questions in which this technique is implemented however I cannot understand the reasoning behind the creation of such a bound for the polynomial. Thank you in advance.

Consider this equality:

-(x+(sqrt(5)))^2 +3

solve for x and you get:

x = sqrt(3)-sqrt(5)

Now, since each second root has a positive and negative answer, this means that there are four solutions because there are four combinations between sqrt(3) and sqrt(5). them being:

-0.5040171699309124028817273272254

3.9681187850686669899366200102371

-3.9681187850686669899366200102371

0.5040171699309124028817273272254

But only the negative solutions work.

Is this related to sqrt(a*b) ≠ sqrt(a)*sqrt(b) assuming a and b are negative?

Thank you.

How do I get numpy.polyfit, to fit to a 3 degree polynomial; so far I've only got the second degree for a curve of points, as I'm only able to change the Y, but not the X based on whether the point has a identification of either -1 or 1 ?

Alright so I've been working on this since 1pm. It's currently 12am and I'm in fucking tears. I want to go to bed but I have no idea what I'm doing thanks to Pearson's shitty teaching.

I've used the "help me solve this" function and got -2 instead of -17. What am I supposed to do?

http://imgur.com/3vIVl8v

Hi all, I'm looking for help understanding why it is that the degree of an expression is the highest degree of any one term, instead of why the degree of an expression isn't the combined sum of the degrees of all terms. Is there a really simple way to understand why this is? I've been trying to find a dumbed down explanation for it, but a lot of what I've come across is over my head or simply, "that's just the way it is". TIA

So I've been normalizing all my inputs/features as (value - mean) / std.

I then turn it into degree 2 polynomial (keeping cross-terms)

I noticed when hand-testing predictions from my model:

• I expect an increase in log-prob when I increase input A
• With a large increase of input A (so squared terms dominate)
• the log-prob would actually decrease!

My issue is that for the squared terms (theres a similar effect for cross-terms), the model sees no difference between a really high score, and a really low score on an input. as low score gets it's minus squared away.

Is this a genuine issue, or something I am imagining, caused by some subtle form of overfitting. i.e. if samples only ever have really low input A for a loss...but never have really really high input A for a win. the large input A squared's will only ever be associated with losses.

also does this sound like an example where using an svm (with non-linear kernel) would be more sensible than logistic regression?

If this is a genuine issue, is there any way I can solve it without moving to degree 3 polynomials (increases the feature number by quite a lot)

I'm haven't taken calculus yet, and I'm wondering if there is a formula that I can create a line that intersects a polynomial at 1 point by taking 2 points from the polynomial (simply for slope), and then translate that slope to a y intercept that only intersects 1?

From my understanding, you have to use the slope to create zeros of multiplicity.

(reworded question incase it didn't make sense: Take 2 points of a polynomial, find their slope. Then can a line be created that intersects the polynomial at 1 point (or double points stacked), with the same slope? If so, what operation can one perform to do this?) (reworded again: a line that intersects tangent to a polynomial, based off of a slope?)

A polynomial P(x) with integer coefficients satisfies the following:

P(5) = 25, P(7) =49, P(9) =81

Find the minimal possible value of |P(10)|

I guessed 100 and that P(x) =x^2. The actual answer is 5. Why? Thanks.

Just like in the title: let R be an integral domain and r its element such that r is prime in R. Consider r as an element of R[X]; is it still prime?

I've tried dabbling in coefficients and some weird induction on polynomial degree, but to no avail. Can anyone point me in the right direction?

I'm talking about points in R^(2), and polynomials f(x,y) in R[x,y], and when I say "passing through", I mean that all of my points are zeros of f(x,y).

The entire graph of f(x,y) = 0 doesn't have to be a simple closed curve, but all of my points have to lie on a single bounded connected component of f(x,y) = 0 and that connected component can't cross itself.

If this is possible, is there a simple construction (analogous to Lagrange interpolation for finding a polynomial y=f(x) through a finite set of points with distinct x-coordinates)?

This is not for anything in particular, I was just wondering. If my question isn't clear, let me know.

) Find a polynomial 𝑞(𝑥) with integer coefficients, and integers 𝑐 and 𝑑 such that c+𝑑∛2 is a root of 𝑞(𝑥), but c − 𝑑∛2 is not a root of 𝑞(𝑥). Describe conditions on 𝑐, 𝑑, and 𝑞(𝑥) for this to hold.

Written in latex:

Let $p(x)$ be a fixed integer polynomial, modulo $m$. Let $N(k)$ be the number of solutions to $p(x)=k$. Show that $\sum_{k=0}^{m-1}N(k)=m$

Let S={1,...,m-1} be a complete residue system mod m. Then for every s in S. f(s)=k, for some k in S and since there are m elements in S. $m\leq \sum_{k=0}^{m-1}N(k)$

My problem is showing that the sum is less then m.

My best attempt is that assuming you order the elements in S from most to least number of solutions. I can get that $\sum_{k=0}^{m-1} N(k)\leq (m^2-\sum_{i=0}^{m-2}\sum_{k=0}^i N(k)$