- #1

mcastillo356

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- TL;DR Summary
- Don't understand Big-O Notation mathematical meaning

Hi, PF

For example, ##\sin{x}=O(x)## as ##x\rightarrow{0}## because ##|\sin{x}|\leq{|x|}## near 0. This fits textbook definition; easy, I think.

But, Taylor's Theorem says that if ##f^{(n+1)}(t)## exists on an interval containing ##a## and ##x##, and if ##P_{n}## is the ##n##th-order Taylor polynomial for ##f## at ##a##, then, as ##x\rightarrow{a}##,

$$f(x)=P_{n}(x)+O\Big({(x-a)^{n+1}}\Big)$$

and "this is a statement about how rapidly the graph of the Taylor polynomial ##P_{n}(x)## aproaches that of ##f(x)## as ##x\rightarrow{a}##; the vertical distance between the graphs

Italic words is what I don't understand

My attempt is somehow to restrict it to one example: MacLaurin polynomial for ##e^x##:

$$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\ldots{+\dfrac{x^n}{n!}+O(x^{n+1})}$$

Why the graph decreases as fast as ##|x|^{n+1}##?

Thanks! Hope LaTeX is right and message is understandable (post without preview, sorry).

For example, ##\sin{x}=O(x)## as ##x\rightarrow{0}## because ##|\sin{x}|\leq{|x|}## near 0. This fits textbook definition; easy, I think.

But, Taylor's Theorem says that if ##f^{(n+1)}(t)## exists on an interval containing ##a## and ##x##, and if ##P_{n}## is the ##n##th-order Taylor polynomial for ##f## at ##a##, then, as ##x\rightarrow{a}##,

$$f(x)=P_{n}(x)+O\Big({(x-a)^{n+1}}\Big)$$

and "this is a statement about how rapidly the graph of the Taylor polynomial ##P_{n}(x)## aproaches that of ##f(x)## as ##x\rightarrow{a}##; the vertical distance between the graphs

*decreases as fast as ##|x-a|^{n+1}##.*Italic words is what I don't understand

My attempt is somehow to restrict it to one example: MacLaurin polynomial for ##e^x##:

$$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\ldots{+\dfrac{x^n}{n!}+O(x^{n+1})}$$

Why the graph decreases as fast as ##|x|^{n+1}##?

Thanks! Hope LaTeX is right and message is understandable (post without preview, sorry).