I need help for this question. I need to find the number of integer solutions for x1 + x2 + x3 + 2x4 + x5 = 68
I'm not sure what to do for 2x4
if you meant:
x^1 + x^2 + x^3 + 2x^4 + x^5 = 68
there are none
http://www.wolframalpha.com/input/?i=solve+x+%2B+x%5E2+%2B+x%5E3+%2B+2x%5E4+%2B+x%5E5+%3D+68
Nope that's not what I meant, it's not to the power. x1 and x2 and so on are there to show different variables.
You can use generating functions. Consider the function
f(y) = sum over x1,...x5 of
y^(x1 + x2 + x3 + 2x4 +x5)
where the summation over x1,...,x5 is from 0 to infinity. This then factors into geometric series, the result is:
f(y) = 1/(1-y)^4 1/(1-y^2) =
-1/(y-1)^5 1/(y+1)
From the definition of f(y), it follows that the number of solutions of x1 + x2 + x3 + 2x4 + x5 = n is the coefficient of y^n in the series expansion of f(y).
To expand f(y) in series in an efficient way, we can write down the partial fractions expansion. Now f(y) has singularities at y = 1 and y = -1, the singular behavior of f(y) at these points yields the partial fraction expansion. This is because the difference between f(y) and the sum of all the singular terms of the expansions around all the singularities of f(y), will have left us with a function without any singularities.
But we started with a rational function and subtracted fractions, so we end up with some rational function, but one without any singularities. Therefore it is in fact a polynomial. But f(y) tends to zero at infinity and so do the terms we subtract from it, so the polynomial is in fact zero. Therefore f(y) is obtained by summing all the singular terms from all the expansions around its singular points.
Near y = -1, we can expand in powers of y+1, the expansion starts with a negative power, 1/(1+y), so this is the only singular term:
f(y) = 1/32 1/(y+1) + ...
Now consider the expansion around y = 1. The singular terms from that expansion plus the term 1/32 1/(1+y) from the expansion around y = -1 equals f(y). But f(y) behaves for large y like y^(-6), this means that the singular terms from the expansion around y = 1 will have to cancel out the leading terms of the large y expansion coming from 1/32 1/(1+y).
Therefore, the smart thing to do is to consider the expand of 1/32 1/(1+y) in powers of (y-1)^(-1) for large y. If we put y = z+1, then we have:
1/32 1/(2+z) =
1/32 1/z 1/(1+ 2/z) =
1/32 1/z [1 - 2/z + 4/z^2 - 8/z^3 + 16/z^4 -...] =
1/32 1/(y-1) - 1/16 1/(y-1)^2 +
1/8 1/(y-1)^3 - 1/4 1/(y-1)^4 +
+1/2 1/(y-1)^5 + ...
The terms of the partial fraction expansion coming from the singularity at y = 1 will have to cancel out all these terms because f(y) decays as y^(-6) for large y, so these partial fraction expansion terms are just minus the above expression (note that we know that it starts with a term proportional to 1/(y-1)^5, so we've captured all the terms). We thus have:
f(y) = 1/32 1/(y+1) - 1/32 1/(y-1) + 1/16 1/(y-1)^2 - 1/8 1/(y-1)^3 +
1/4 1/(y-1)^4 - 1/2 1/(y-1)^5
which we can rewrite as:
f(y) = 1/32 1/(y+1) + 1/32 1/(1-y) + 1/16 1/(1-y)^2 + 1/8 1/(1-y)^3 +
1/4 1/(1-y)^4 + 1/2 1/(1-y)^5
To expand this in powers of y, you can use that the coefficient of y^r in 1/(1-y)^p is Binomial(r+p-1,p-1). So the coefficient of y^68 is:
1/32 [2 + 2*69 + 4*70*69/2 + 8*71*70*69/6 + 16*72*71*70*69/24] =
528990
To find the number of integer solutions for the given equation, we need to use a technique called "stars and bars" or "balls and urns."
Let's first ignore the term 2x4 and focus on the equation x1 + x2 + x3 + x5 = 68. This is a standard equation that can be solved using stars and bars.
To illustrate this technique, imagine having 68 stars (representing the sum of the variables) and four bars (representing the four variables x1, x2, x3, x5). We need to distribute the stars between these four bars in order to find the solutions.
Now, let's determine the number of ways to arrange these 68 stars and four bars. Each arrangement represents a unique solution to the equation.
To find the number of arrangements, we can think of it as a permutation problem. We have a total of 72 objects (68 stars and 4 bars), and we need to arrange them. The number of arrangements can be calculated using the formula for permutations with repeated elements:
N = (n + r - 1)C(r - 1),
where N is the number of arrangements, n is the total number of objects (stars + bars), and r is the number of distinct objects (here, the number of bars).
Substituting the values, we get:
N = (68 + 4 - 1)C(4 - 1),
N = 71C3.
Now, let's consider the term 2x4. Since 2x4 must be even, it can take only even values. We can divide the equation into two cases: when 2x4 = 0 and when 2x4 = 2.
Case 1: When 2x4 = 0, we consider it as a separate variable. So the equation becomes:
x1 + x2 + x3 + x5 + 0 = 68,
x1 + x2 + x3 + x5 = 68,
which we have already solved using stars and bars.
Case 2: When 2x4 = 2, we divide both sides of the equation by 2 to get:
x1 + x2 + x3 + x5 + 1 = 68/2,
x1 + x2 + x3 + x5 = 34.
Again, we can solve this using stars and bars.
By calculating the number of arrangements in Case 1 and Case 2 separately and summing them up, we can find the total number of integer solutions for x1 + x2 + x3 + 2x4 + x5 = 68.
In summary, to find the number of integer solutions for x1 + x2 + x3 + 2x4 + x5 = 68, you need to use the stars and bars technique for cases 2x4 = 0 and 2x4 = 2 separately and then add the number of arrangements from both cases.