Is it correct to say that terms in a taylor series expansion

>>8576379
>Is it correct to say that terms in a taylor series expansion are the first differential, second differential, third... etc.?
they are the n-th differential at expansion point x0.

>>8576381

When we integrate, we use first differential quantities, dx. Is there calculus that integrates with second differentials?

>>8576385
Just intégrate twice :^)

>>8576394

That doesnt make sense. If use an integral operator on every term in an expanded taylor series what would that equal to?

>>8576394

Is there a calculus text that centers around the topics of integration and differentiation with taylor series expansion?

>>8576398
Taylor series in 0 :
[eqn]f(x) = f(0) + \frac{x}{1!}*f'(0) + \frac{x^2}{2!}*f''(0) + ... [/eqn]
All the way to infinity.
So you integrate term-to-term.
[eqn]F(x) = 0 + \frac{x^2}{2*1!}*f'(0) +\frac{x^3}{3*2!}*f''(0) + ... [/eqn]
All the way to infinity. So basically, while it has a nice patters, it doesn't really amount to anything.

>>8576477
My bad. I made a big mistake.
[eqn]F(x) = Constant + x + \frac{x^2}{2*1!}*f'(0) +\frac{x^3}{3*2!}*f''(0) + ...[/eqn]

And since [math]f''(0) = F'''(0)[/math] :

[eqn]F(x) = Constant + x + \frac{x^2}{2*1!}*F''(0) +\frac{x^3}{3*2!}*F'''(0) + ...[/eqn].

So basically when you integrate, you get the Taylor series of the primitive. Kinda logical when you think about it (fix constant to F(0).)

>>8576477

Want to rethink evaluating the first term there buddy?

>>8576491
Yes, see >>8576488

I'm also pretty sure you can find a constant where

[math]Constant + x = F(0) + F'(0)*x[/math]

>>8576488
>>8576491

Nice. Lol

Any book suggestions that focuses calculus around taylor series?