Let's denote the three numbers by a1, a2, and a3. Consider the third degree polynomial:
p(x) = (1 + a1 x)(1 + a2 x)(1 + a3 x)
Take the logarithm:
Log[p(x)] =
Log(1 + a1 x) + Log(1 + a2 x)
+ Log(1 + a3 x)
Expand in powers of x by using that:
Log(1 + x) = x - x^2/2 +x^3/3 - x^4/4 + ...
Then, if we denote the sum of the n-th powers of the ai by Sn, we find:
Log[p(x)] = S1 - S2/2 x^2 + S3/3x^3 - S4/4 x^4 + ...
If we now exponentiate this we should get p(x) back:
Exp{Log[p(x)]} = p(x)
We can calculate
Exp[S1 - S2/2 x^2 + S3/3x^3 - S4/4 x^4 + ...]
using the series expansion:
Exp[X] = 1 + X + X^2/2! + X^3/3! + X^4/4! + ...
If we put in here X = S1 - S2/2 x^2 + S3/3x^3 - S4/4 x^4 + ...]
the result must be p(x). Snce p(x) is a third degree polynomial, this means that
all powers of x^4 and higher must vanish.
If you work out the coefficient of x^4 and equate it to zero you get the equation:
S4 = S1^4/6 - S1^2 S2 + S2^2/2 + 4/3 S1 S3
So, if S1 = 4, S2 = 10, and S3 = 22, then S4 must be 50.
the sum of three numbers is 4, the sum of their squares is 10 and the sum of their cubes is 22. what is the sum of their fourth powers?
To solve this problem, we need to find the sum of the fourth powers of the three numbers given that the sum, sum of squares, and sum of cubes are known.
Let's call the three numbers a, b, and c.
Given that the sum of the three numbers is 4, we have:
a + b + c = 4 ----(1)
Given that the sum of their squares is 10, we have:
a^2 + b^2 + c^2 = 10 ----(2)
Given that the sum of their cubes is 22, we have:
a^3 + b^3 + c^3 = 22 ----(3)
We want to find the sum of their fourth powers, which can be denoted as:
a^4 + b^4 + c^4 = S4
To solve for S4, we can use the equation derived from the expansion of the logarithm and exponentiation as mentioned in the question:
S4 = S1^4/6 - S1^2 S2 + S2^2/2 + 4/3 S1 S3
Here, S1, S2, and S3 represent the sum, sum of squares, and sum of cubes respectively.
Substituting the given values of S1 = 4, S2 = 10, and S3 = 22, we get:
S4 = 4^4/6 - 4^2 * 10 + 10^2/2 + 4/3 * 4 * 22
Simplifying the equation, we get:
S4 = 64/6 - 160 + 100/2 + 352/3
S4 = 10.6666 - 160 + 50 + 117.3333
S4 = 18 + 50 + 117.3333
S4 = 185.3333
So, the sum of the fourth powers of the three numbers is approximately 185.3333.
To find the sum of the fourth powers of the three numbers, let's denote the three numbers as a, b, and c.
Given that the sum of the numbers is 4:
a + b + c = 4 ...(1)
And the sum of their squares is 10:
a^2 + b^2 + c^2 = 10 ...(2)
And the sum of their cubes is 22:
a^3 + b^3 + c^3 = 22 ...(3)
Now, we need to find the sum of their fourth powers, which is denoted as S4.
According to the given information, let's calculate S4 using the equation derived in the question:
S4 = S1^4/6 - S1^2 S2 + S2^2/2 + (4/3) S1 S3
Substituting the values of S1 = a + b + c = 4, S2 = a^2 + b^2 + c^2 = 10, and S3 = a^3 + b^3 + c^3 = 22, we can find S4:
S4 = (4)^4/6 - (4)^2 (10) + (10)^2/2 + (4/3)(4)(22)
= 64/6 - 160 + 100/2 + (4/3)(4)(22)
= 10.6667 - 160 + 50 + 35.3333
= -64
Therefore, the sum of the fourth powers of the three numbers is -64.