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
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
3 answers
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
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
3 answers