(x+y)^4 can be expanded starting at x^4*y^0 and taking the sum of the products of the derivative of the x term and the integral of the y term? Why does this work?
Expanding (x+y)^4 using the method you described does not yield the correct result. The expansion of (x+y)^4 is obtained by applying the Binomial Theorem, which states that:
(x+y)^n = nC0 * x^n * y^0 + nC1 * x^(n-1) * y^1 + nC2 * x^(n-2) * y^2 + ... + nC(n-1) * x^1 * y^(n-1) + nCn * x^0 * y^n
Here, nCk represents "n choose k," which is used to calculate combinations. nCk = n! / (k! * (n-k)!)
To explain why the method you mentioned does not work, we need to consider the concept of derivatives and integrals. In calculus, a derivative measures the rate of change of a function at a given point, while an integral calculates the area under a curve. However, in the expansion of (x+y)^4, we are not dealing with derivatives or integrals.
Expanding the expression using the Binomial Theorem works because it accounts for all possible combinations of powers of x and y that can arise from expanding an expression with a binomial base raised to a positive integer power.
To simplify the expansion of (x+y)^4, we can substitute n = 4 in the Binomial Theorem:
(x+y)^4 = 4C0 * x^4 * y^0 + 4C1 * x^3 * y^1 + 4C2 * x^2 * y^2 + 4C3 * x^1 * y^3 + 4C4 * x^0 * y^4
Simplifying further, we get:
(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4
So, the correct expansion of (x+y)^4 does not involve derivatives or integrals, but rather the coefficients of the binomial coefficients (nCk) and the appropriate powers of x and y.