Can the constant term of a power series be zero?

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SUMMARY

The discussion centers on the conditions under which the constant term of a power series can be zero, particularly in the context of linear differential equations. It is established that for the initial value problem defined by the equation y'' + y = 0 with initial conditions y(0) = 0 and y'(0) = 1, the constant term cannot be zero as it leads to a trivial solution. The Frobenius method is highlighted, demonstrating that if the coefficient a_0 is zero, it alters the solution technique for ordinary differential equations (ODEs). The conversation ultimately reveals that the distinction between the constant term and the first non-zero term in a series is a matter of semantics.

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  • Understanding of linear differential equations and their solutions
  • Familiarity with power series and their representations
  • Knowledge of the Frobenius method for solving ODEs
  • Basic concepts of initial value problems in differential equations
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gikiian
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In the context of my work (linear differential equations), it can not be zero. But why?
 
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gikiian said:
In the context of my work (linear differential equations), it can not be zero. But why?

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?
 
y'' + y = 0, y(0) = 0, y'(0) = 1.
 
I get it! It will satisfy the equation, but the solution will be a trivial one.
 
There is only one solution for the initial value problem I wrote in post #3, and it is not the trivial solution.
 
gikiian said:
In the context of my work (linear differential equations), it can not be zero. But why?
What about, say, a function ##f:\mathbb{R}\to\mathbb{R}, x\mapsto x##?
 
gikiian said:
Can the constant term of a power series be zero?
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!}+...

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

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.
 

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