When does x=a in the taylor series stop being x=a?

[krobben]

I'm having a hard time understanding the fundamentals of the taylor series. So I get how you continually take derivatives in order to find the coefficients but in order to do that we have to state that x=a. Well when we finally get done we have an infinite polynomial of f(x)=f(a)+f'(a)(x-a)/1!+f''(a)(x-a)/2!..… So this is where I get completely lost. We got to this point only under the circumstance that x=a(unless I'm missing something), so if x=a then how do we have f(x)=f(a)+f'(a)(x-a)/!+f''(a)(x-a)/2!... without x=a because I'm under the impressions that this could only end up being f(x)=f(a)+f'(a)(0)/1!+f''(a)(0)/2!... since x=a thus x-a = a-a and so x-a =0. Is there something I'm not understanding or is there a point when we abandon x=a and if so why can we do that?

 

[SammyS]

I'm having a hard time understanding the fundamentals of the taylor series. So I get how you continually take derivatives in order to find the coefficients but in order to do that we have to state that x=a. Well when we finally get done we have an infinite polynomial of f(x)=f(a)+f'(a)(x-a)/1!+f''(a)(x-a)/2!..… So this is where I get completely lost. We got to this point only under the circumstance that x=a(unless I'm missing something), so if x=a then how do we have f(x)=f(a)+f'(a)(x-a)/!+f''(a)(x-a)/2!... without x=a because I'm under the impressions that this could only end up being f(x)=f(a)+f'(a)(0)/1!+f''(a)(0)/2!... since x=a thus x-a = a-a and so x-a =0. Is there something I'm not understanding or is there a point when we abandon x=a and if so why can we do that?

Indeed, f(a)=f(a)+f'(a)(0)/1!+f''(a)(0)/2!...

That's f(a) on the left hand side, not f(x)

 

[DryRun]

You could compare it to Maclaurin series where x=0

 

[krobben]

I really do appreciate your answers and help all together but I'm still confused here. I'm curious as to how f(x)=f(a)+f'(a)(x-a)/!+f''(a)(x-a)/2!... is possible as a variable equation when we have already declared that x must equal a. Obviously this isn't true but I want to know why we can stop saying x must equal a and that x is now released as a variable.

 

[Mark44]

I really do appreciate your answers and help all together but I'm still confused here. I'm curious as to how f(x)=f(a)+f'(a)(x-a)/!+f''(a)(x-a)/2!... is possible as a variable equation when we have already declared that x must equal a.

Nowhere does it say that x must equal a. I don't know where you're getting this idea.

If x = a, then both sides of the equation above simplify to f(a).

Obviously this isn't true but I want to know why we can stop saying x must equal a and that x is now released as a variable.

 

[HallsofIvy]

I'm having a hard time understanding the fundamentals of the taylor series. So I get how you continually take derivatives in order to find the coefficients but in order to do that we have to state that x=a. Well when we finally get done we have an infinite polynomial of f(x)=f(a)+f'(a)(x-a)/1!+f''(a)(x-a)/2!..… So this is where I get completely lost. We got to this point only under the circumstance that x=a(unless I'm missing something), so if x=a then how do we have f(x)=f(a)+f'(a)(x-a)/!+f''(a)(x-a)/2!... without x=a because I'm under the impressions that this could only end up being f(x)=f(a)+f'(a)(0)/1!+f''(a)(0)/2!... since x=a thus x-a = a-a and so x-a =0. Is there something I'm not understanding or is there a point when we abandon x=a and if so why can we do that?

One of the things you are missing is the powers on (x- a)! You should
have f(x)= f(a)+ f'(a)(x- a)/1!+ f''(a)(x-a)^2/2!+ f'''(a)(x-a)^3/3!+ ...
is the "Taylor series of f about the point x= a".

To take a very simple example, look at the polynomial $f(x)= x^3- 3x^2+ 3x- 1$. Then $f'(x)= 3x^2- 6x+ 3$, $f''(x)= 6x- 6$, $f'''(x)= 6$ and all succeeding derivatives are 0.

To find the Taylor's series about x= 2, note that f(2)= 8- 12+ 6- 1= 1, f'(2)= 12- 12+ 3= 3, f''(2)= 12- 6= 6, and f'''(2)= 6 so that the Taylor's series is [itex]6+ 3(x- 2)/1!+ 6(x- 2)^2/2!+ 6(x- 2)^3/3!= 6+ 3(x- 2)+ 3(x- 2)^2+ (x- 2)^3[itex]. <br /> <br /> To find the Taylor's series about x= 3, note that f(3)= 27- 27+ 9- 1= 8, f'(3)= 27- 27+ 9- 1= 8, f''(3)= 18- 6= 12, and f'''(3)= 6 so that the Taylor's series is [itex]8+ 8(x- 3)/1!+ 12(x- 3)^2/2!+ 6(x- 3)^3/3!= 6+ 8(x- 3)+ 6(x- 3)^2+ (x- 3)^3[itex]. <br /> <br /> To find the Taylor's series about x= 0, note that f(0)= -1, f'(0)= 3, f''(0)= -6, and f'''(0)= 6 so that the Taylor's series is [itex]-1+ 3(x- 0)/1!- 6(x- 0)^2/2!+ 6(x- 0)^3/3!= -1+ 3x+ 3x^2+ x^3$. Of course, that is exactly the polynomial we started with. In fact, if you multiply out the first two, you will see that they also give exactly the same thing.<br /> <br /> By the way, it is NOT always true that a function is always equal to its Taylor's series. That is true for all "analytic" functions (which are very nice functions and typically what you see in Calculus) but not true for <b>all<b> infinitely differentiable functions.</b></b>$[/itex][/itex][/itex][/itex][/itex]

 

[DryRun]

OK, i think i understand what the OP is confused about:

f(x)=a is not the same as x=a.

Also, when the problem states that x=2, this does not mean that you need to replace "x" by "2" in the Taylor polynomial. I know it sounds confusing, but when the problem says "about x=2" then it just means about the point "2" and by that, you should understand that it means "a=2".

After you've replaced "a=2" in the Taylor polynomial, the answer will be in terms of the variable x, hence another indication why you cannot eliminate "x" by replacing it with a constant in the polynomial.

 

[Ray Vickson]

I'm having a hard time understanding the fundamentals of the taylor series. So I get how you continually take derivatives in order to find the coefficients but in order to do that we have to state that x=a. Well when we finally get done we have an infinite polynomial of f(x)=f(a)+f'(a)(x-a)/1!+f''(a)(x-a)/2!..… So this is where I get completely lost. We got to this point only under the circumstance that x=a(unless I'm missing something), so if x=a then how do we have f(x)=f(a)+f'(a)(x-a)/!+f''(a)(x-a)/2!... without x=a because I'm under the impressions that this could only end up being f(x)=f(a)+f'(a)(0)/1!+f''(a)(0)/2!... since x=a thus x-a = a-a and so x-a =0. Is there something I'm not understanding or is there a point when we abandon x=a and if so why can we do that?

It might help to look at what is meant by f'(a), f''(a), etc. We have
$$f'(a) = \lim_{h \rightarrow 0}\frac{f(a+h) - f(a)}{h} ,$$ and
$$f''(a) = \lim_{h \rightarrow 0} \frac{f(a+h) + f(a-h) - 2f(a)}{h^2}.$$ (You may not know this last one, but it holds for any sufficiently smooth function f.) Alternatiely, we can put
$$f''(a) = \lim_{h \rightarrow 0} \frac{f'(a+h) - f'(a)}{h},$$
where f' is the function given by
$$f'(t) = \lim_{s \rightarrow 0} \frac{f(t+s)-f(t)}{s}$$ for any t. Again, there is no 'x' here. Part of your confusion may stem from trying to use the symbol 'x' in two different ways in the same formula. When you realize that when defining the derivatives, 'x' just acts as a dummy variable, so can be replaced by some other variable t or w or whatever.

RGV