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Mathematics 19 Online
OpenStudy (anonymous):

Which of the following are true statements? All Taylor series are power series. Taylor polynomials and Taylor series are the same thing. No Taylor function exists that exactly equals f(x) = sin(x).

OpenStudy (anonymous):

I only II only I and III only I and II only

ganeshie8 (ganeshie8):

Taylor series is a special case of power series with \[a_n=\dfrac{f^n(c)}{n!}\]

OpenStudy (dan815):

1 is right 2 is i guess| 3 debatable

ganeshie8 (ganeshie8):

I'm inclining toward I and III only

OpenStudy (dan815):

like if u need an infinite number of terms to get sinx.. then is that really saying it exists

ganeshie8 (ganeshie8):

For any finite \(n\), the remainder is never \(0\), so \(T_n(x)\) never identically equals \(\sin(x)\)

OpenStudy (dan815):

ya id go with i and iii only too

OpenStudy (anonymous):

\[T_n(x) = \sum_{i=0}^{n} \frac{ f^{(i)}(a) }{ i! }(x-a)^i\] Tn is a polynomial of degree n, so I don't think II is right

OpenStudy (anonymous):

We can also use f(x) = e^x and check the result

OpenStudy (anonymous):

That's a bit insulting ganeshie! Haha, but it's funny! XD

ganeshie8 (ganeshie8):

i better delete it before kids come to this thread

OpenStudy (anonymous):

Haha good idea

OpenStudy (anonymous):

For anyone that was wondering, I picked I and III and it said it is wrong. Now 3 of these answers have I in it, so either there are some Taylor series that aren;t power series (but that is not true, Taylor is just a subset) or there is a difference between the taylor series and the taylor polynomials

ganeshie8 (ganeshie8):

whats the definition of "taylor function" that your professor is using ?

OpenStudy (anonymous):

Idk. The people that make the tests are a different company than the lessons (this is online)

ganeshie8 (ganeshie8):

Okay the question depends on the definitions we use for : 1) taylor series (clear) 2) taylor polynomial (almost clear) 3) taylor function (a bit ambiguous)

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