When to stop to be sure of the pattern of signs in a Taylor expansion?

Question

It is common practice to derive just "enough" terms of a series expansion to illustrate the pattern. For example, upon seeing some terms one can be certain how the negative sign appears alternatingly or how the coefficient relates to the power.

$$\log(1 + x) = x - \frac{x^2}2 + \frac{ x^3 }3 - \frac{ x^4 }4 + \cdots \\ \tan^{-1} x = x - \frac{x^3}3 + \frac{ x^5 }5 - \frac{ x^7 }7 + \cdots $$

How does one know when is "enough" when deriving the series?

Some series appears to have the signs go like $++--++--\cdots$ (perhaps with zeros interleaved). How does one know the pattern truly is "alternating by two"?

Suppose the series starts with the first four terms of plus signs $++++$, how does one know if it's all plus sign or if it's $++++----++++----\cdots$?

^{Note 1: my question is clearly different from this post with a similar title.}

^{Note 2: I'm not sure how my question can be addressed by the Descartes rule of signs.}

Answer 1

To be blunt, it's typically not a matter of noticing a pattern, but proving the pattern that shows when it's enough.

That's not to say pattern recognition is not helpful, because it absolutely is. It can cue you into where proving the pattern might lead, or what you might have to prove - all depending on the context. But the statement

$$\tan^{-1} x = x - \frac{x^3}3 + \frac{ x^5 }5 - \frac{ x^7 }7 + \cdots = \sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1} x^{2k+1}$$

means you would need to show that the factor controlling the sign, $(-1)^k$, does indeed show up in the summation. You can't just say "oh I found the first few terms, the rest obviously follow the same pattern." I mean, they might, but there are examples of sequences that suddenly break from such a pattern. Maybe yours breaks from the pattern $1,000,$ or perhaps $100,000,000$, or perhaps $10^{{10}^{100}}$ terms down the road.

I mean, you have the right idea, sort of, you almost seem to be on the brink of realizing this very thing at the end of your post - that after finding however-many terms of the sequences, how do you know that the pattern doesn't change? And that's the thing - that's completely true, there's no reason, just finding

$$x \;\;\; , \;\;\; - \frac{x^3}3 \;\;\; , \;\;\; \frac{ x^5 }5 \;\;\; , \;\;\; - \frac{ x^7 }7 \;\;\; , \;\;\; ...$$

that you should expect $x^9/9$ to come next. Finding a pattern isn't sufficient, you also have to verify - to prove - that pattern. I believe this post has a rough proof for the arctangent power series.

Answer 2

Well, do recall where the Taylor expansion comes from.

$$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2+\cdots$$

Note the derivatives in each term of the expansion. This makes proving some patterns easy - for example, we know that the derivative of $\sin x$ cycles every 4 derivatives.

For one of your examples, $\log(1+x)$, note that the first derivative is $\frac{1}{1+x}$ and then it is pretty easy to show by the power rule for derivatives that the sign alternates.

For the other example, $\tan^{-1} x$, note that the Taylor series for this function is derived from the fact that the derivative is $\frac{1}{1+x^2}$ and rewriting that as a geometric series with common ratio $-x^2$. It's easy to see that the alternating sign goes on forever, since it is a geometric series.

You can apply this reasoning to most common Taylor series expansions.

Answer 3

To avoid the difficulty and questionable accuracy of guesswork, direct computation is always a good route. I will walk you step-by step through the derivation of each series you mentioned.

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First one: $\log(1+x)$

Recall that for $|t|<1$, $$\frac1{1+t}=\sum_{n\geq0}(-1)^nt^n$$ integrating both sides from $0$ to $x$, with $|x|<1$, $$\log(1+x)=\sum_{n\geq1}(-1)^{n-1}\frac{x^{n}}n$$

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Next one $\arctan x$:

Recall that $$\frac{\mathrm d}{\mathrm dt}\arctan t=\frac1{1+t^2}$$ So for $|t|<1$, $$\frac{\mathrm d}{\mathrm dt}\arctan t=\sum_{n\geq0}(-1)^nt^{2n}$$ Applying $\int_0^x$ on both sides, $$\arctan x=\sum_{n\geq0}(-1)^n\frac{x^{2n+1}}{2n+1}$$ Which converges for $|x|<1$.

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This technique is not just useful for finding Taylor Series (because it avoids the whole $n$-th derivative thing), but it's also helpful for finding closed forms for series.

Example: The evaluation of $$S=\sum_{n\geq1}\frac{3^n}{n{2n\choose n}}$$

First we recall that $${2n\choose n}^{-1}=\frac{(n!)^2}{(2n)!}=\frac{n}2\frac{((n-1)!)^2}{(2n-1)!}$$ Then recall the definition of the Beta function $$\mathrm{B}(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}\mathrm dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ Where $\Gamma(s)=(s-1)!$ is known as the Gamma function.

Anyway we may now observe that $${2n\choose n}^{-1}=\frac{n}2\mathrm{B}(n,n)$$ Next consider the series $$S(x)=\sum_{n\geq1}\frac{x^n}{n{2n\choose n}}$$ We may now observe that $$S(x)=\frac12\sum_{n\geq1}x^n\mathrm{B}(n,n)$$ $$S(x)=\frac12\sum_{n\geq1}x^n\int_0^1 [t(1-t)]^{n-1}\mathrm dt$$ $$S(x)=\frac12\int_0^1\sum_{n\geq1}x^n[t(1-t)]^{n-1}\mathrm dt$$ $$S(x)=\frac12\int_0^1x\sum_{n\geq0}[xt(1-t)]^{n}\mathrm dt$$ Then using the Geometric series, $$S(x)=\frac{x}2\int_0^1\frac{\mathrm dt}{1-xt(1-t)}$$ $$S(x)=\frac{x}2\int_0^1\frac{\mathrm dt}{xt^2-xt+1}$$ For this integral, complete the square in the denominator and preform a trigonometric substitution to see that for $0<x<4$, $$S(x)=2\sqrt{\frac{x}{4-x}}\arctan\sqrt{\frac{x}{4-x}}$$ Then plugging in $x=3$ gives $$S(3)=S=2\sqrt{3}\arctan\sqrt{3}$$ And since $\arctan\sqrt3=\pi/3$, $$S=\frac{2\pi}{\sqrt3}$$

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I guess the lesson here is that when doing this sort of thing, start with what you know, then go to where you don't know, instead of trying to do something crazy like finding a general expression for $\frac{\mathrm d^n}{\mathrm dt^n}\arctan t\big|_{t=0}$ or tryng to guess a general expression.