Calculus/Calculus 2


Taylor Series
Taylor Series  
Please help me understand this problem with a helpful step-by-step explanation. I really do not even know where to start. I recognize this is a Taylor series, but aside from that, I am completely lost. Thank you so much for you help in advance.

~All necessary info should be in the picture.~

Sorry about the delay, but I haven't been able to do this for a few days.  Here's my answer, though ...

This is what a Taylor's series does.  It approximates a continuous function that is continuously differentiable at some point by using a nearby point where the function can be evaluated.

To describe it, the point we take is x0.  We then evaluate the function and many derivatives at that point.  To determined the value at some point x, it is given by ...
let me say that f(n)(x0) is the nth derivative evaluated at x0,
(x-x0)^n is the nth power of (x-x0), and
n! is n factorial.

This is the series: T(x) = [f(n)(x0)][(x-x0)^n]/[n!].

To write out the first few terms, let's use x0=1.
The first few terms in the factorial are 0!, 1!, 2!, 3!, and 4!.
The values are 0! = 1, 1! = 1*0! = 1, 2!= 2*1! = 2; 3! = 3*2! = 6; and 4! = 4*3! = 4*6 = 24.

They Taylor's function would then be
T(x) = f(1)/1 + f'(1)(x-1)/1 + f"(x)(x-1)/2 + f"'(x)(x-1)/6 + f""(x)[(x-1)^4]/24.

Note that the brackets are just to make sure it is understood that x-1 is raised to the 4th.
For the terms with a 2 and a 3, the are characters { alt 253 } and { I don't even remember what that one is.  That's why I have them in a spreadsheet of mathematical terms so I can just highlight them, cut them, and paste them in here. }.

Also note that f"'(x0) is the 3rd derivative evaluated at x0 and
f""(x0) is the 4th derivative evaluated at x0.

All that needs to be known is the basic function.  That is, evaluate the nth derivative at some point x0, compute, for the point we're interested in, which is x, (x-x0)^n {nth power of (x-x0) } and compute n!.  The factorial m! is just the product of the first n integers.
There is a rather complicated expression that gives 0! = 1, but it is best to just know that.
Note the terms I gave were 1, 1, 2,6, 24, but they continue on.  5! = 5*4! = 5*24 = 120;
6! = 6*5! = 6*120 = 720; etc.  They get big fast, and they're in the denominator, so the fractions make progressive terms fade out quickly.


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