Math

Question Find a function ff such that an=f(n)a_{n}=f(n) for recurrence relations: an+2=an+1+an+n2,a1=1,a2=2a_{n+2}=a_{n+1}+a_{n}+n^{2}, a_{1}=1, a_{2}=2 or an+2=4an+1+4an,a1=1,a2=2a_{n+2}=4 a_{n+1}+4 a_{n}, a_{1}=1, a_{2}=2.

Studdy Solution

STEP 1

Assumptions1. We are given two recurrence relations and initial conditions. . We need to find a function f(n)f(n) such that an=f(n)a_{n}=f(n) for each recurrence relation.

STEP 2

Let's start with the first recurrence relation an+2=an+1+an+n2,a1=1,a2=2a_{n+2}=a_{n+1}+a_{n}+n^{2}, a_{1}=1, a_{2}=2.

STEP 3

We will solve this recurrence relation by using the method of generating functions. A generating function is a series that encodes the terms of a sequence.

STEP 4

Let's define the generating function (x)(x) for the sequence ana_{n} as follows(x)=a1x+a2x2+a3x3+a4x4+(x) = a_{1}x + a_{2}x^{2} + a_{3}x^{3} + a_{4}x^{4} + \cdots

STEP 5

We can rewrite the recurrence relation in terms of (x)(x)(x)a1xa2x2=x(x)a1x+x2(x)a2x2+x33+x44+(x) - a_{1}x - a_{2}x^{2} = x(x) - a_{1}x + x^{2}(x) - a_{2}x^{2} + \frac{x^{3}}{3} + \frac{x^{4}}{4} + \cdots

STEP 6

implify the above equation to get(x)(1xx2)=a1x+a2x2x33x44(x)(1 - x - x^{2}) = a_{1}x + a_{2}x^{2} - \frac{x^{3}}{3} - \frac{x^{4}}{4} - \cdots

STEP 7

olve the above equation for (x)(x)(x)=a1x+a2x2x33x441xx2(x) = \frac{a_{1}x + a_{2}x^{2} - \frac{x^{3}}{3} - \frac{x^{4}}{4} - \cdots}{1 - x - x^{2}}

STEP 8

The above equation gives the generating function for the sequence ana_{n}. To find the function f(n)f(n), we need to find the coefficient of xnx^{n} in the series expansion of (x)(x).

STEP 9

Unfortunately, finding the explicit form of f(n)f(n) from the generating function is not straightforward and often requires special techniques or cannot be done at all. In this case, we cannot find a simple explicit form for f(n)f(n).

STEP 10

Now, let's move on to the second recurrence relation an+2=4an++4an,a=,a2=2a_{n+2}=4 a_{n+}+4 a_{n}, a_{}=, a_{2}=2.

STEP 11

Again, we will use the method of generating functions. Define the generating function G(x)G(x) for the sequence ana_{n} as followsG(x)=ax+ax+a3x3+a4x4+G(x) = a_{}x + a_{}x^{} + a_{3}x^{3} + a_{4}x^{4} + \cdots

STEP 12

We can rewrite the recurrence relation in terms of G(x)G(x)G(x)axa2x2=4xG(x)4ax+4x2G(x)4a2x2G(x) - a_{}x - a_{2}x^{2} =4xG(x) -4a_{}x +4x^{2}G(x) -4a_{2}x^{2}

STEP 13

implify the above equation to getG(x)(xx2)=ax+a2x2G(x)( -x -x^{2}) = a_{}x + a_{2}x^{2}

STEP 14

olve the above equation for G(x)G(x)G(x)=ax+a2x24x4x2G(x) = \frac{a_{}x + a_{2}x^{2}}{ -4x -4x^{2}}

STEP 15

The above equation gives the generating function for the sequence ana_{n}. To find the function f(n)f(n), we need to find the coefficient of xnx^{n} in the series expansion of G(x)G(x).

STEP 16

Again, finding the explicit form of f(n)f(n) from the generating function is not straightforward. In this case, we cannot find a simple explicit form for f(n)f(n).
In conclusion, while we can express the given recurrence relations in terms of generating functions, finding explicit forms for the functions f(n)f(n) is not straightforward and often cannot be done at all.

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