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So as I understand so far: A power series is like any other series except now the partial sums depend on the variable x. The value of x determines the convergence or divergence of the series, meaning at certain x values the nth partial sum goes to infinity, and at other x values the nth partial sum actually goes to a number. There is an interval over which any x value within that interval will cause the series to converge. Here is what I do not understand:

So over this interval the power series represents a function of x, but what happens outside of that interval?

The sum of the series goes to infinity, but does it no longer approximate the function?

Couldn't it approximate the same function as it goes to infinity?

Thank you very much to anyone willing to help me

Rdewolfe
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2 Answers2

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In complex analysis (calculus and analysis with complex valued variables) you will learn that a convergent power series represents what's called a holomorphic "complex differentiable" function, and that the power series can often be transformed so that it is valid and equal to the same homomorphic function in a region beyond the radius of convergence of the original series. Also, I don't remember the details, but there are examples in complex analysis of divergent series that can still give a good approximation for the solution to a problem if you choose the right cutoff term before the series diverges too much.

user2566092
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You have correctly described the key points about power series.

Let the function you are concerned with be $f(x)$

Suppose you have obtained a power series $P(x)=a_0+a_1x+a_2x^2+a_3x^3+...$

Define the partial power series $P_n(x)=a_0+a_1x+a_2x^2+a_3x^3+...a_nx^n$

For some values of $x$, we find that the values $P_1$, $P_2$, $P_3$, ... converge to a limit. If this limit is equal to $f(x)$, then we have a power series for $f(x)$ that is valid for those values of $x$.

However for other values of $x$, we find that the values $P_1$, $P_2$, $P_3$, ... do not converge. We then say that the power series is not valid for those values of $x$.

For example, it can be shown using the binomial expansion that $\frac 1 {1+x} = 1 - x+x^2-x^3+x^4-x^5+...$

Setting $x=0.1$ gives $1 - 0.1+0.01-0.001+0.0001-0.00001+...$

The values that you get if you take 1, 2, 3 and more terms of this expression are: 1, 0.9, 0.91, 0.909, 0.9091, 0.90909, ...

They are converging to the correct value of $\frac 1 {1+ 0.1}=\frac 1 {1.1} = \frac {10} {11}$

BUT setting $x=2$ gives $1 - 2+4-8+16-32+...$

The values that you get if you take 1, 2, 3 and more terms of this expression are: 1, -1, 3, -5, 11, -21, ...

These values are not converging to the correct value of $\frac 1 {1+ 2} = \frac 1 3$

We say that the power series is valid for $x=0.1$ and not valid for $x=2$. In fact we can identify in this case that the power series is valid if $\left| x \right|<1$

tomi
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