Math Is Fun Forum

baegelion

Member

Registered: 2016-09-14

Posts: 4

Functions

1. Let F(x) be the real-valued function defined for all real x except for x = 0 and x = 1 and satisfying the functional equation F(x) + F((x-1)/x) = 1+x. Find the F(x) satisfying these conditions.

Write F(x) as a rational function with expanded polynomials in the numerator and denominator.

2. Suppose that f(x) and g(x) are functions which satisfy f(g(x)) = x^2 and g(f(x)) = x^3 for all x ≥ 1. If g(16) = 16, then compute log_2 g(4). f(x) ≥ 1 and g(x) ≥ 1 for all x ≥ 1

3. The function

satisfies xf(x) + f(1 - x) = x^3 - x for all real x. Find f(x).

baegelion

Member

Registered: 2016-09-14

Posts: 4

Re: Functions

4. Suppose we have the following identity:

Find the minimum of

over 0 ≤ p ≤ 1.

5. What is the maximum degree of a polynomial of the form

with

for 0 ≤ i ≤ n, 1 ≤ n, such that all the zeros are real?

6. Let f(m,1) = f(1,n) = 1 for m ≥ 1, n ≥ 1, and let f(m,n) = f(m-1,n) + f(m,n-1) + f(m-1,n-1) for m > 1 and n > 1. Also, let

, for a ≥ 1, b ≥ 1.
Note: The summation notation means to sum over all positive integers a,b such that a+b=k.
Given that
S(k+2) = pS(k+1) + qS(k) for all k ≥ 2,
for some constants p and q, find pq.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Functions

Hi;

---

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.

baegelion

Member

Registered: 2016-09-14

Posts: 4

Re: Functions

#4 is not 1.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Functions

Hi;

---

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.

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Functions

5. What is the maximum degree of a polynomial of the form

with

for 0 ≤ i ≤ n, 1 ≤ n, such that all the zeros are real?

Some clarification is required about

since

just the second degree expression has complex zero.

---

{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Functions

Last edited by thickhead (2016-09-17 03:46:07)

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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

baegelion

Member

Registered: 2016-09-14

Posts: 4

Re: Functions

5. What is the maximum degree of a polynomial of the form

with

for 0 ≤ i ≤ n, 1 ≤ n, such that all the zeros are real?

Some clarification is required about

since

just the second degree expression has complex zero.

x^2 - x - 1 also fits the form and has two positive zeroes.

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Functions

x^2 - x - 1 also fits the form and has two positive zeroes.

You mean 2 real roots.It means one has to select whatever fits.

---

{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Functions

Hi baegelion;

5)

has real roots.

---

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.

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Functions

(5) maximum degree is 3. I tried with 4th degree with  all + and - variations  (total 16 forms) in wolfram alpha and all have  complex roots.

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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}