# Can the constant term of a power series be zero?

1. Nov 28, 2013

### gikiian

In the context of my work (linear differential equations), it can not be zero. But why?

2. Nov 28, 2013

### SteamKing

Staff Emeritus
It's not clear why your context implies that the constant term cannot equal zero. If the constant term were zero, would the series not satisfy the differential equation and any associated initial conditions?

3. Nov 28, 2013

### D H

Staff Emeritus
y'' + y = 0, y(0) = 0, y'(0) = 1.

4. Nov 28, 2013

### gikiian

I get it! It will satisfy the equation, but the solution will be a trivial one.

5. Nov 28, 2013

### D H

Staff Emeritus
There is only one solution for the initial value problem I wrote in post #3, and it is not the trivial solution.

6. Nov 28, 2013

### Mandelbroth

What about, say, a function $f:\mathbb{R}\to\mathbb{R}, x\mapsto x$?

7. Nov 29, 2013

### gikiian

Yes. Take $xe^{x}$ for example:
$xe^{x}$
$=x(1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+...)$
$=x+x^{2}+\frac{x^{3}}{2!}+\frac{x^{4}}{3!}+\frac{x^{5}}{4!}+...$
$=0+x+x^{2}+\frac{x^{3}}{2!}+\frac{x^{4}}{3!}+\frac{x^{5}}{4!}+...$

In Frobenius method, you have $y=\sum^{∞}_{n=0} a_{n}x^{n+r}=x^{r}(a_{o}+a_{1}x+a_{2}x^{2}+...)$. Now if $a_{o}$ becomes 0, then the series would become $y=x^{r}(0+a_{1}x+a_{2}x^{2}+...)=x^{r}(a_{1}x+a_{2}x^{2}+...)=x^{r}x(a_{1}+a_{2}x+a_{3}x^{2}+...)=x^{r+1}(a_{1}+a_{2}x+a_{3}x^{2}+...)$. This would essentially change the mathematical technique that we use here to solve the ODE. Hence we 'assume' that $a_{o}$ can not be 0.

Last edited: Nov 29, 2013
8. Nov 29, 2013

### AlephZero

OK, now we can see this is a question about semantics, not series solutions of differential equations.

"The constant term of a power series" means $c_0$ in the series $c_0 + c_1x + c_2x^2 + \dots$.

But in post #7, you are just saying that every non-trivial series must have first non-zero term, and you are calling that term $a_0$. it is the coefficient of $x^r$. It is not the constant term of the series solution unless $r = 0$.

9. Nov 29, 2013

### gikiian

I am a little confused as to what you are saying here, and would like to understand your point better. Please explain what are you trying to convey.