Is there a way how to improve this way of expanding polynomials?

Question

I was just playing around and came up with this 'method of expanding polynomials'. I start with $$(x+a)^2$$, the $a$ being a constant, which I differentiate to obtain $$2(x+a)$$ I multiply this out to obtain $$2x+2a$$ and integrate term by term to get $$ x^2+2ax+C $$ where I am losing a constant.

Now if I didn't want to lose the constant I would consider $(x+y)^2$ , or if I wanted to compute $(x+y+z)^2$, I would have to differentiate these expressions. The problem is that if I differentiate with respect to $x$, I lose all terms that depend only on $y$ (and $z$).

Is there a way to overcome this obstacle?

Answer 1

You have to find the constant of integration for our specific case. Specifically, you know that when $x = 0$, we have $(x+a)^2 = a^2$. That means that $a^2 = 0^2 + 2\cdot 0a + C = C$, so we get $C = a^2$.

ALternatively, you could use that when $x = -a$, we have $(x + a)^2 = 0$, which means that $0 = (-a)^2 + 2(-a)a + C = -a^2 + C$, again giving $C = a^2$.

Or, you could use any other value for $x$. They all tell you that $C = a^2$. The main point is that you have a funciton $2x + 2a$, and to this function you have the general antiderivative $x^2 + 2xa + C$, and you have a specific antiderivative $(x + a)^2$. In order to figure out which calue for $C$ gives you the specific antiderivative you're after, you need to evaluate the two somewhere and compare their values.

Answer 2

You can also differentiate wrt to the other variables; so differentiating $(x+y)^2$ wrt first to $x$ and then $y$ you get $$2x + 2y$$ in both cases .Then integrate first wrt to $x$ and then to $y$ to get

$$x^2 + 2xy + C(y)$$ $$y^2 + 2xz + D(x)$$

where $C$, $D$ are unknown functions. But the two expression must be equal so you can write

$$x^2 + C(y) = y^2 + D(x) \implies x^2 - D(x) = y^2 - C(y)$$

the LHS depends solely on $x$ and the RHS solely on $y$, so they both have to be equal to a constant $k$; from which you find

$$D(x) = x^2 - k$$ and $$C(y) = y^2 - k$$

set $x=y=0$ in the original expression to recover $k = 0$.

Answer 3

Actually, the following is a broken by a single constant:

$$x=\int\left(\frac d{dx}x\right)dx=\int1dx=x+C$$

Indeed, this is why the FTOC is usually formatted as follows:

$$f(x)=\frac d{dx}\int_a^xf(t)dt$$

with the derivative taken after the integral.