# Function problems

How to solve those polynomial functions problems on brilliant? ex.f(x)=2f(x+1)+(x+5)...something.

Note by Michael Martin
5 years, 9 months ago

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You will need to gain familiarity with what polynomials are, and various ways that you can manipulate them. There are several basics you should know about polynomials.
1. The degree of a polynomial
2. Factorization
3. Remainder-factor theorem
4. Calculating the coefficients of a polynomial given several initial values
5. Rational Root Theorem, Integer Root Theorem
6. Determining zeroes by looking at $$f(a) > 0, f(b) < 0$$ and applying the Intermediate Value Theorem.
7. Conjugate roots
8.

A. Lagrange Interpolation Formula
B. Understanding roots in greater detail
C. If $$f$$ is a polynomial with integer coefficients, then $$a - b | f(a) - f(b)$$
D. Irreducibility Criterion - Gauss lemma, Eisenstein criterion
E. Special polynomials like Symmetric, Cyclic, Orthogonal, Chebyshev, etc.
F. Integrating and differentating a polynomial
G.

Anyone else has suggestions to add on to this list? That's why 7, 8, F, G are empty. List them below and I'd add them on.

Staff - 5 years, 9 months ago

What does the vertical bar in C mean?

- 5 years, 9 months ago

- 5 years, 9 months ago

Any special book recommendations?

- 5 years, 9 months ago

I don't exactly know if there are better methods of doing this, but what I did last week sort of looks like the procedure for decomposing into partial fractions.

Let $$f(x)=Ax^{n}+Bx^{n-1}+...$$ where $$n$$ depends on the problem (usually the highest degree of $$x$$ in the given equation). Then solve for each term in the given equation, manipulate as necessary, and group similar terms in $$x$$.

You should obtain something like $$(3A)x+(2A-B)=-5x+2$$ (or of course something more complicated). Equate the coefficients of terms that have similar literal coefficients, and solve the resulting system of equations. Substitute back into $$f(x)$$, and there it is.

I hope my explanation is clear, I'm kind of in a hurry because I have a deadline to meet. :)

- 5 years, 9 months ago

How to solve questions like: Find all functions such that f(x + yf(x)) = x.f(x) + f(y) and many more? HELP ME...

- 3 years, 8 months ago

it depends on its type whether constant or linear or quadratic function

- 5 years, 9 months ago

You can write this function f(x) ,as a polynomial if you know the value for someone.Like this Suppose that f(0) is equal to 1.So you will have the value for f(1) and f(2)... But for an expression non recursive,eu do this. f(x+1)=2f(x+2)+x+1 + 5. if we make f(x+1)-f(x) we will have f(x+1)-f(x)=2f(x+2)+x +6 -2f(x+1)-x-5 So you gonna have an recursive expression without the term of 1°degree.Then you do the "process" again and you wil have an recursive expression without a terma constant.And now,you can solve this equation.

- 5 years, 9 months ago

By determining degree of f(x) you reduce your work to almost 3rd grader stuff !!!!! For example last week a problem was as follows: Find f(10) if $$f(x)+(x+1)^{3}=2f(x+1)$$ So by observation we know that f(x) can never be a polynomial of any other degree except 3, else there would not be any balance of $$x^3$$ on the left and right hand side .......

- 5 years, 9 months ago

Your reasoning of "else there would not be any balance" is not entirely correct.

Staff - 5 years, 9 months ago

- 5 years, 9 months ago

I know for $$x \neg 0$$and for my example the reasoning is sound. Could you elaborate on your comment that my logic isn't entirly correct.I know I wrote it in haste but for my understanding , please do elaborate !!!

- 5 years, 9 months ago

To determine the degree of $$f(x)$$ in last week's problem, we may proceed as follows (let us denote this degree by a positive integer $$n$$):

If $$n<3$$, then the right-hand side of the equation will not contain a cubic term. However, the left-hand side will have a cubic term from $$(x+1)^3$$, and hence the equation cannot possibly hold.

Too bad I'm sleepy, I can't remember how I eliminated $$n \geq 4$$. I lost my solution, and I'm working as I type this anyway. LOL but I hope I got it right so far. Sorry for not finishing this.

- 5 years, 9 months ago

I'm not entirely sure if by 'balance of $$x^3$$' you mean that the coefficients on both side is non-zero (that's my assumption), or that the net coefficient is 0 (in which case, my comment is invalid.

If the degree of the polynomial is $$\geq 3$$, then there can be various $$x^3$$ terms on both sides. So, your statement only applies for $$n = 0, 1, 2$$.

Staff - 5 years, 9 months ago

Yes..... what I meant was that if I were to choose any polynomial than a 3rd degree, On the LHS I would have Lets say a as the coefficient of $$x^n$$, whereas on the RHS I would have 2a as the coefficient for $$x^n$$ ..... so ultimately the coefficients of all powers of x higher than 3 will have a coefficient equal to zero.

- 5 years, 9 months ago