Understanding Taylor Series: Finding the General Formula | Math Explained"

[PFuser1232]

$$f(a + x) = \sum_{k=0}^∞ \frac{f^{(k)}(a) x^k}{k!}$$

Usually written as:

$$f(t) = \sum_{k=0}^∞ \frac{f^{(k)}(a) (t-a)^k}{k!}$$

Where $t = a + x$
Is the taylor expansion supposed to give the same result for all $a$? The reason this confuses me is because this seems to suggest that $f(1 + x) = f(4 + x) = f(π + x)$ and so on, which is usually not the case. Where did I go wrong?

 

[Stephen Tashi]

Is the taylor expansion supposed to give the same result for all $a$?

Yes, in the sense that for a given t

f(t) = \sum_{k=0}^\infty \frac{f^{(k)}(a_1) (t-a_1)^k}{k!} = \sum_{k=0}^\infty \frac{f^{(k)}(a_2)(t-a_2)^k}{k!}

when both series converge.

this seems to suggest that $f(1 + x) = f(4 + x) = f(π + x)$ and so on, which is usually not the case. Where did I go wrong?

In the first version you listed for Taylor series, to say that the equation holds for different a_1, a_2 does not imply f(a_1 + x) = f(a_2 + x) since, in general, f^{(k)}(a_1) \ne f^{(k)}(a_2). In the second version you listed, different values of a_1, a_2 do not imply different values of t.

 

[PFuser1232]

Yes, in the sense that for a given t

f(t) = \sum_{k=0}^\infty \frac{f^{(k)}(a_1) (t-a_1)^k}{k!} = \sum_{k=0}^\infty \frac{f^{(k)}(a_2)(t-a_2)^k}{k!}

when both series converge.

In the first version you listed for Taylor series, to say that the equation holds for different a_1, a_2 does not imply f(a_1 + x) = f(a_2 + x) since, in general, f^{(k)}(a_1) \ne f^{(k)}(a_2). In the second version you listed, different values of a_1, a_2 do not imply different values of t.

So the variable $x$ must also change for $t$ to remain well-defined.

 

[Stephen Tashi]

So the variable $x$ must also change for $t$ to remain well-defined.

The meaning of that statement isn't entirely clear, but I'm tempted to say yes.

 

[PFuser1232]

The meaning of that statement isn't entirely clear, but I'm tempted to say yes.

In order to maintain the same meaning of $t$ (and $f(t)$) while changing the value of $a$, we must change the value of $x$.

 

[Stephen Tashi]

In order to maintain the same meaning of $t$ (and $f(t)$) while changing the value of $a$, we must change the value of $x$.

Yes, if we assert t = a + x.

 

[PFuser1232]

Yes, if we assert t = a + x.

If we were to swap the positions of all $a$'s and $x$'s in the expansion; that is: $f(x) + f'(x)a + \frac{f''(x)a^2}{2!} + ...$, would we get the same result? I understand that our goal is to write a power series where the powers are constantly increasing on a variable, not a constant. But I'm curious, would this give the same result?

 

[Stephen Tashi]

If we were to swap the positions of all $a$'s and $x$'s in the expansion; that is: $f(x) + f'(x)a + \frac{f''(x)a^2}{2!} + ...$, would we get the same result? I understand that our goal is to write a power series where the powers are constantly increasing on a variable, not a constant. But I'm curious, would this give the same result?

A logician could object to calling such a change in the formula "the same result" since logician wants it made clear which symbols are variables (in some scope) and which are constants (in some scope). Technically, a symbol like "x", should appear with the scope of a quantifier (like "for each") when we write a mathematical statement. People who do calculus are careful about quantifying their variables when they do epsilon-delta proofs (e.g. for each epsilon > 0, there exists a delta ...such that...), but in other aspects of calculus they are careless. You can talk about "Taylor's Formula" as a string of symbols. To ask a precise mathematical question about its meaning you need to first state "Taylor's Theorem" as a theorem. See which quantifiers apply to which variables. Then you can ask whether the symbols representing the variables can be swapped.

 

[KevinMWHM]

(a_n)_1^infty

 

[KevinMWHM]

({a_n})_1^ \infty