Solved on Oct 30, 2023

Find a function $f$ such that $a_{n}=f(n)$ for recurrence relations: $a_{n+2}=a_{n+1}+a_{n}+n^{2}, a_{1}=1, a_{2}=2$ or $a_{n+2}=4 a_{n+1}+4 a_{n}, a_{1}=1, a_{2}=2$ .

STEP 1

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

STEP 2

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

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 $(x)$ for the sequence $a_{n}$ as follows $(x) = a_{1}x + a_{2}x^{2} + a_{3}x^{3} + a_{4}x^{4} + \cdots$

STEP 5

We can rewrite the recurrence relation in terms of $(x)$ $(x) - a_{1}x - a_{2}x^{2} = x(x) - a_{1}x + x^{2}(x) - a_{2}x^{2} + \frac{x^{3}}{3} + \frac{x^{4}}{4} + \cdots$

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

Unfortunately, finding the explicit form of $f(n)$ 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 $f(n)$ .

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

implify the above equation to get $G(x)( -x -x^{2}) = a_{}x + a_{2}x^{2}$

STEP 14

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

STEP 15

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

STEP 16

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

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Find a function $f$ such that $a_{n}=f(n)$ for recurrence relations: $a_{n+2}=a_{n+1}+a_{n}+n^{2}, a_{1}=1, a_{2}=2$ or $a_{n+2}=4 a_{n+1}+4 a_{n}, a_{1}=1, a_{2}=2$ .

STEP 1

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

STEP 2

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

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 $(x)$ for the sequence $a_{n}$ as follows $(x) = a_{1}x + a_{2}x^{2} + a_{3}x^{3} + a_{4}x^{4} + \cdots$

STEP 5

We can rewrite the recurrence relation in terms of $(x)$ $(x) - a_{1}x - a_{2}x^{2} = x(x) - a_{1}x + x^{2}(x) - a_{2}x^{2} + \frac{x^{3}}{3} + \frac{x^{4}}{4} + \cdots$

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

Unfortunately, finding the explicit form of $f(n)$ 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 $f(n)$ .

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

implify the above equation to get $G(x)( -x -x^{2}) = a_{}x + a_{2}x^{2}$

STEP 14

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

STEP 15

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

STEP 16

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STEP 1

Assumptions1. We are given two recurrence relations and initial conditions. . We need to find a function f(n)f(n)f(n) such that an=f(n)a_{n}=f(n)an​=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}=2an+2​=an+1​+an​+n2,a1​=1,a2​=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)(x) for the sequence ana_{n}an​ as follows(x)=a1x+a2x2+a3x3+a4x4+⋯(x) = a_{1}x + a_{2}x^{2} + a_{3}x^{3} + a_{4}x^{4} + \cdots(x)=a1​x+a2​x2+a3​x3+a4​x4+⋯

STEP 5

We can rewrite the recurrence relation in terms of (x)(x)(x)(x)−a1x−a2x2=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(x)−a1​x−a2​x2=x(x)−a1​x+x2(x)−a2​x2+3x3​+4x4​+⋯

STEP 6

implify the above equation to get(x)(1−x−x2)=a1x+a2x2−x33−x44−⋯(x)(1 - x - x^{2}) = a_{1}x + a_{2}x^{2} - \frac{x^{3}}{3} - \frac{x^{4}}{4} - \cdots(x)(1−x−x2)=a1​x+a2​x2−3x3​−4x4​−⋯

STEP 7

olve the above equation for (x)(x)(x)(x)=a1x+a2x2−x33−x44−⋯1−x−x2(x) = \frac{a_{1}x + a_{2}x^{2} - \frac{x^{3}}{3} - \frac{x^{4}}{4} - \cdots}{1 - x - x^{2}}(x)=1−x−x2a1​x+a2​x2−3x3​−4x4​−⋯​

STEP 8

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

STEP 9

Unfortunately, finding the explicit form of f(n)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)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}=2an+2​=4an+​+4an​,a​=,a2​=2.

STEP 11

Again, we will use the method of generating functions. Define the generating function G(x)G(x)G(x) for the sequence ana_{n}an​ as followsG(x)=ax+ax+a3x3+a4x4+⋯G(x) = a_{}x + a_{}x^{} + a_{3}x^{3} + a_{4}x^{4} + \cdotsG(x)=a​x+a​x+a3​x3+a4​x4+⋯

STEP 12

We can rewrite the recurrence relation in terms of G(x)G(x)G(x)G(x)−ax−a2x2=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}G(x)−a​x−a2​x2=4xG(x)−4a​x+4x2G(x)−4a2​x2

STEP 13

implify the above equation to getG(x)(−x−x2)=ax+a2x2G(x)( -x -x^{2}) = a_{}x + a_{2}x^{2}G(x)(−x−x2)=a​x+a2​x2

STEP 14

olve the above equation for G(x)G(x)G(x)G(x)=ax+a2x2−4x−4x2G(x) = \frac{a_{}x + a_{2}x^{2}}{ -4x -4x^{2}}G(x)=−4x−4x2a​x+a2​x2​

STEP 15

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

STEP 16

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

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

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

We can rewrite the recurrence relation in terms of $(x)$ $(x) - a_{1}x - a_{2}x^{2} = x(x) - a_{1}x + x^{2}(x) - a_{2}x^{2} + \frac{x^{3}}{3} + \frac{x^{4}}{4} + \cdots$

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

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

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

Unfortunately, finding the explicit form of $f(n)$ 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 $f(n)$ .

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

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

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

implify the above equation to get $G(x)( -x -x^{2}) = a_{}x + a_{2}x^{2}$

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

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