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There is no solution to the problem.

Proof:

We have to consider two cases:

Case a) at least one of the $P_j$ is zero

The let k be the smallest index for which $P_k = 0$.

Then from

$0 = \alpha P_k = P_{k+1} + ... + P_{2k+1}$

we have

$P_{k+1} = P_{k+2} = .. = P_{2k+1} = 0$

and, inductively, $P_j = 0$ if $j \ge k$.

On the other hand

$\alpha P_{k-1} = P_{k} + ... = 0$

and so on downwards so that all $P_j = 0$ which contradicts the normalization condition. Hence we can rule out case a)

Case b) all P_j are positive

I shall show that there is no solution to the recursive equations with

(1) $P_j \gt 0, j = 1, 2, 3, ...$

First from

$\alpha P_1 = P_2 + P_3$

we conclude

$\alpha \gt 0$

Notice also the $P_0$ appears only in the relation

$\alpha P_0 = P_1$

which shows that

$P_0 \gt 0$ as well, but $P_0$ does not appear in the normalization. Therefore we consider it as a mere abrevíation for $P_1/ \alpha $.

Now we transform the recursive relation into a standard form, which we define here to be one in which an element with a specific index is defined in terms of elements with smaller indices.

Define

(2) $Q_i = P_{i+1}+P_{i+2}+..., i = 0,1,2,...$

As a sum over positive quantities we have

$Q_i \gt 0, i = 0, 1, 2, ...$

The inversion of (2) is

(3) $P_i = Q_{i-1} - Q_i , i = 1, 2, ... $

Now the equations become

$\alpha P_j = Q_j - Q_{2j+1}, j =1, 2, 3, ... $

Using (3) we get

$\alpha (Q_{j-1}-Q_j) = Q_j - Q_{2j+1}$

or

(4) $Q_{2j+1} = (1+\alpha ) Q_j - \alpha Q_{j-1}, j = 1, 2, ...$

This is now a recursive relation in standard form.

The inital values are

$Q_0 = P_1 + P_2 + ... = 1$

because of the normalization condition.

And

$Q_1 = 1 - P_1 = 1 - \alpha P_0$

can be considered as a free parameter in the interval (0,1).

Before we solve (4) we observe that it defines only the elements with an odd index.

Therefore we let

$Q_{2k} = C_k > 0, k = 1, 2, ...$

with arbitrary $C_k$ in the interval (0,1).

Performing now the first few steps of the solution to (4) the reader will find that

$P_{10} = - C_5 - \alpha (1+\alpha ) Q_1$

But this is a negative quantity, and the contradiction proves the statement.

There is no solution to the problem in the first version of the OP.

Proof:

We have to consider two cases:

Case a) at least one of the $P_j$ is zero

The let $k$ be the smallest index for which $P_k = 0$.

Then from

$0 = \alpha P_k = P_{k+1} + ... + P_{2k+1}$

we have

$P_{k+1} = P_{k+2} = .. = P_{2k+1} = 0$

and, inductively, $P_j = 0$ if $j \ge k$.

On the other hand

$\alpha P_{k-1} = P_{k} + ... = 0$

and so on downwards so that all $P_j = 0$ which contradicts the normalization condition. Hence we can rule out case a)

Case b) all $P_j$ are positive

I shall show that there is no solution to the recursive equations with

(1) $P_j \gt 0, j = 1, 2, 3, ...$

First from

$\alpha P_1 = P_2 + P_3$

we conclude

$\alpha \gt 0$

Notice also the $P_0$ appears only in the relation

$\alpha P_0 = P_1$

which shows that

$P_0 \gt 0$ as well, but $P_0$ does not appear in the normalization. Therefore we consider it as a mere abrevíation for $P_1/ \alpha $.

Now we transform the recursive relation into a standard form, which we define here to be one in which an element with a specific index is defined in terms of elements with smaller indices.

Define

(2) $Q_i = P_{i+1}+P_{i+2}+..., i = 0,1,2,...$

As a sum over positive quantities we have

$Q_i \gt 0, i = 0, 1, 2, ...$

The inversion of (2) is

(3) $P_i = Q_{i-1} - Q_i , i = 1, 2, ... $

Now the equations become

$\alpha P_j = Q_j - Q_{2j+1}, j =1, 2, 3, ... $

Using (3) we get

$\alpha (Q_{j-1}-Q_j) = Q_j - Q_{2j+1}$

or

(4) $Q_{2j+1} = (1+\alpha ) Q_j - \alpha Q_{j-1}, j = 1, 2, ...$

This is now a recursive relation in standard form.

The inital values are

$Q_0 = P_1 + P_2 + ... = 1$

because of the normalization condition.

And

$Q_1 = 1 - P_1 = 1 - \alpha P_0$

can be considered as a free parameter in the interval (0,1).

Before we solve (4) we observe that it defines only the elements with an odd index.

Therefore we let

$Q_{2k} = C_k > 0, k = 1, 2, ...$

with arbitrary $C_k$ in the interval (0,1).

Performing now the first few steps of the solution to (4) the reader will find that

$P_{10} = - C_5 - \alpha (1+\alpha ) Q_1$

But this is a negative quantity, and the contradiction proves the statement.

I shall show that there is no solution to the recursive equations with

(1) $P_j \gt 0, j = 1, 2, 3, ...$

First from

$\alpha P_1 = P_2 + P_3$

we conclude

$\alpha \gt 0$

Notice also the $P_0$ appears only in the relation

$\alpha P_0 = P_1$

which shows that

$P_0 \gt 0$ as well, but $P_0$ does not appear in the normalization. Therefore we consider it as a mere abrevíation for $P_1/ \alpha $.

Now we transform the recursive relation into a standard form, which we define here to be one in which an element with a specific index is defined in terms of elements with smaller indices.

Define

(2) $Q_i = P_{i+1}+P_{i+2}+..., i = 0,1,2,...$

As a sum over positive quantities we have

$Q_i \gt 0, i = 0, 1, 2, ...$

The inversion of (2) is

(3) $P_i = Q_{i-1} - Q_i , i = 1, 2, ... $

Now the equations become

$\alpha P_j = Q_j - Q_{2j+1}, j =1, 2, 3, ... $

Using (3) we get

$\alpha (Q_{j-1}-Q_j) = Q_j - Q_{2j+1}$

or

(4) $Q_{2j+1} = (1+\alpha ) Q_j - \alpha Q_{j-1}, j = 1, 2, ...$

This is now a recursive relation in standard form.

The inital values are

$Q_0 = P_1 + P_2 + ... = 1$

because of the normalization condition.

And

$Q_1 = 1 - P_1 = 1 - \alpha P_0$

can be considered as a free parameter in the interval (0,1).

Before we solve (4) we observe that it defines only the elements with an odd index.

Therefore we let

$Q_{2k} = C_k > 0, k = 1, 2, ...$

with arbitrary $C_k$ in the interval (0,1).

Performing now the first few steps of the solution to (4) the reader will find that

$P_{10} = - C_5 - \alpha (1+\alpha ) Q_1$

But this is a negative quantity, and the contradiction proves the statement.

There is no solution to the problem.

Proof:

We have to consider two cases:

Case a) at least one of the $P_j$ is zero

The let k be the smallest index for which $P_k = 0$.

Then from

$0 = \alpha P_k = P_{k+1} + ... + P_{2k+1}$

we have

$P_{k+1} = P_{k+2} = .. = P_{2k+1} = 0$

and, inductively, $P_j = 0$ if $j \ge k$.

On the other hand

$\alpha P_{k-1} = P_{k} + ... = 0$

and so on downwards so that all $P_j = 0$ which contradicts the normalization condition. Hence we can rule out case a)

Case b) all P_j are positive

I shall show that there is no solution to the recursive equations with

(1) $P_j \gt 0, j = 1, 2, 3, ...$

First from

$\alpha P_1 = P_2 + P_3$

we conclude

$\alpha \gt 0$

Notice also the $P_0$ appears only in the relation

$\alpha P_0 = P_1$

which shows that

$P_0 \gt 0$ as well, but $P_0$ does not appear in the normalization. Therefore we consider it as a mere abrevíation for $P_1/ \alpha $.

Now we transform the recursive relation into a standard form, which we define here to be one in which an element with a specific index is defined in terms of elements with smaller indices.

Define

(2) $Q_i = P_{i+1}+P_{i+2}+..., i = 0,1,2,...$

As a sum over positive quantities we have

$Q_i \gt 0, i = 0, 1, 2, ...$

The inversion of (2) is

(3) $P_i = Q_{i-1} - Q_i , i = 1, 2, ... $

Now the equations become

$\alpha P_j = Q_j - Q_{2j+1}, j =1, 2, 3, ... $

Using (3) we get

$\alpha (Q_{j-1}-Q_j) = Q_j - Q_{2j+1}$

or

(4) $Q_{2j+1} = (1+\alpha ) Q_j - \alpha Q_{j-1}, j = 1, 2, ...$

This is now a recursive relation in standard form.

The inital values are

$Q_0 = P_1 + P_2 + ... = 1$

because of the normalization condition.

And

$Q_1 = 1 - P_1 = 1 - \alpha P_0$

can be considered as a free parameter in the interval (0,1).

Before we solve (4) we observe that it defines only the elements with an odd index.

Therefore we let

$Q_{2k} = C_k > 0, k = 1, 2, ...$

with arbitrary $C_k$ in the interval (0,1).

Performing now the first few steps of the solution to (4) the reader will find that

$P_{10} = - C_5 - \alpha (1+\alpha ) Q_1$

But this is a negative quantity, and the contradiction proves the statement.

I shall show now that there is no solution to the recursive equations with

(1) $P_j \gt 0, j = 1, 2, 3, ...$

First from

$\alpha P_1 = P_2 + P_3$

we conclude

$\alpha \gt 0$

Notice also the $P_0$ appears only in the relation

$\alpha P_0 = P_1$

which shows that

$P_0 \gt 0$ as well, but $P_0$ does not appear in the normalization. Therefore we consider it as a mere abrevíation for $P_1/ \alpha $.

Now we transform the recursive relation into a standard form, which we define here to be one in which an element with a specific index is defined in terms of elements with smaller indices.

Define

(2) $Q_i = P_{i+1}+P_{i+2}+..., i = 0,1,2,...$

As a sum over positive quantities we have

$Q_i \gt 0, i = 0, 1, 2, ...$

The inversion of (2) is

(3) $P_i = Q_{i-1} - Q_i , i = 1, 2, ... $

Now the equations become

$\alpha P_j = Q_j - Q_{2j+1}, j =1, 2, 3, ... $

Using (3) we get

$\alpha (Q_{j-1}-Q_j) = Q_j - Q_{2j+1}$

or

(4) $Q_{2j+1} = (1+\alpha ) Q_j - \alpha Q_{j-1}, j = 1, 2, ...$

This is now a recursive relation in standard form.

The inital values are

$Q_0 = P_1 + P_2 + ... = 1$

because of the normalization condition.

And

$Q_1 = 1 - P_1 = 1 - \alpha P_0$

can be considered as a free parameter in the interval (0,1).

Before we solve (4) we observe that it defines only the elements with an odd index.

Therefore we let

$Q_{2k} = C_k > 0, k = 1, 2, ...$

with positive $C_k$

Performing now the first few steps of the solution to (4) the Reader will find that

$P_{10} = - C_5 - \alpha (1+\alpha ) Q_1$

But this is a negative quantity, and the contradiction proves the statement.

I shall show that there is no solution to the recursive equations with

(1) $P_j \gt 0, j = 1, 2, 3, ...$

First from

$\alpha P_1 = P_2 + P_3$

we conclude

$\alpha \gt 0$

Notice also the $P_0$ appears only in the relation

$\alpha P_0 = P_1$

which shows that

$P_0 \gt 0$ as well, but $P_0$ does not appear in the normalization. Therefore we consider it as a mere abrevíation for $P_1/ \alpha $.

Now we transform the recursive relation into a standard form, which we define here to be one in which an element with a specific index is defined in terms of elements with smaller indices.

Define

(2) $Q_i = P_{i+1}+P_{i+2}+..., i = 0,1,2,...$

As a sum over positive quantities we have

$Q_i \gt 0, i = 0, 1, 2, ...$

The inversion of (2) is

(3) $P_i = Q_{i-1} - Q_i , i = 1, 2, ... $

Now the equations become

$\alpha P_j = Q_j - Q_{2j+1}, j =1, 2, 3, ... $

Using (3) we get

$\alpha (Q_{j-1}-Q_j) = Q_j - Q_{2j+1}$

or

(4) $Q_{2j+1} = (1+\alpha ) Q_j - \alpha Q_{j-1}, j = 1, 2, ...$

This is now a recursive relation in standard form.

The inital values are

$Q_0 = P_1 + P_2 + ... = 1$

because of the normalization condition.

And

$Q_1 = 1 - P_1 = 1 - \alpha P_0$

can be considered as a free parameter in the interval (0,1).

Before we solve (4) we observe that it defines only the elements with an odd index.

Therefore we let

$Q_{2k} = C_k > 0, k = 1, 2, ...$

with arbitrary $C_k$ in the interval (0,1).

Performing now the first few steps of the solution to (4) the reader will find that

$P_{10} = - C_5 - \alpha (1+\alpha ) Q_1$

But this is a negative quantity, and the contradiction proves the statement.