Question

Would someone give me good questions in taylor series/mclaurins? Thanks

Answer 1

Find the Taylor series about 2 for f(x) = x^-1.

Answer 2

What is the Taylor series expansion of
f(x) = ln x about x = 1?

Answer 3

i cant answer estudier's problem. hahaha

Answer 4

Just so we are all clear, a Taylor series about 0 = a Maclaurin series.

So people often refer to just a Maclaurin series when the center is 0.

Answer 5

estudier,please give me a hint!

Answer 6

You need to calculate f(2), (f'(2), f''(2) and so on.

Answer 7

i did that! but the answer is 3+4(x-2) +2(x-2)^2 / 2! +0!

Answer 8

f(2) = 1/2, f'(2) = -x^(-2) = -1/(2^2)

etc.

Answer 9

how do you make the answer into like this \[\sum_{\infty}^{i=0}\]

Answer 10

aw lol my bad. i thought it's x^2 - 1 haha

Answer 11

Just find the pattern...

Answer 12

could you solve x^2 - 1? i cant get the answer :|

Answer 13

I don't understand the question.

Answer 14

Oh. Could you get the mclaurins expansion for f(x)= x^2 - 1

Answer 15

oh i meant taylor series of expansion. x=2

Answer 16

Can I ask a question?

Are you interested in this subject generally or are you more interested in approximations?

Answer 17

Subject. Sorry for making you spoonfed me.

Answer 18

So you have read the definition of a Taylor polynomial (about 0)?

Answer 19

yep. it's just that i have a test coming up and it pressures me. it's called mclaurins, right?

Answer 20

A Mclaurin series is the same as a Taylor series about 0.

But I asked you about a Taylor polynomial.

Answer 21

Oh sorry! we're not there yet. we only have tackled mclaurins and taylor series of expansion and operations.

Answer 22

Polynomials usually are dealt with first as an introduction to Taylor series.

Answer 23

Oh. But our instructor did not teach Taylor polynomial. He just taught us how to solve mclaurins and taylor series.

Answer 24

But not the ones that I and abtrehearn provided?

Answer 25

I already answered them. I only followed what my instructor told me to do. -.-

Answer 26

The difference between a Taylor polynomial and a Taylor series is that the first only requires an n-times fifferentiable function whereas the other requires an infinitely differentiable function.

Answer 27

Oh. So we're dealing with infinitely diff. functions. Thanks estudier for explaining!

Answer 28

That is in the definition.

Answer 29

Your x^2 -1 can only be differentiated twice.

Answer 30

I really need to familiarize imyself with this by means of practicing. But there are still many basic probs that i cant solve.

Answer 31

So you will get a linear approximation to the quadratic.

Answer 32

but how should i make it into this form:
\[\sum_{\infty}^{i=0} f^{i} (x-a)^{i} /i! \]

Answer 33

oh my algebraic sum is wrong :p

Answer 34

You find the pattern.

Answer 35

but there's only 3 terms and it ends with +0 and there's no pattern in f^i(a)

Answer 36

That's a Taylor polynomial, not a series- you can't sum to infinity!

Answer 37

oh sorry i forgot thanks for mentioning again!

Answer 38

hahaha now i understand it alot better

Answer 39

It seems to me that you have been studying Taylor polynomials not Taylor series.

Answer 40

well i made that question up thinking that it's taylor series haha

Answer 41

Well, it is in a way, just truncated.

Answer 42

I would say that it is a Taylor series with a finite number of terms.

Answer 43

i meant taylor infinity series. thanks again estudier! you're a great help!

Answer 44

REALLY! :)

Answer 45

If eg you take (1+x)^4 and sub in the binomial series,you will get the expansion (a Taylor series) 1 + 4x + 6x^2 etc. (valid for x between 1 and -1).

Answer 46

It's OK just to learn the methods up to a point but eventually it causes a bit of a problem if you don't understand at least some of the definitions and theory.

Answer 47

Well at least now, i have the advantage thanks to you! hey can you give me more problems in taylor series (infinity)?

Answer 48

You haven't answered the two you were given yet:-)

Answer 49

Do you want some simpler ones?

Answer 50

i answered it already! the one you gave is
1/2 + \[\sum_{\infty}^{i=1} (-1)^{i} (x-2)^{i} / 2^{i}\]

Answer 51

and for the ln x

Answer 52

Sorry...and its 1/2 - not 1/2 +

Answer 53

\[\ln 10 + \sum_{i=0}^{\infty} (-1)^{i} (x-10)^{i} / 10^{i+1} i\]

Answer 54

waaaaaaaa!

Answer 55

but i started with the negative sign in -1^i which is i=1

Answer 56

Taylor series about 0 for 1/(3+x)^2 and specify the range of validity.

Answer 57

range of validity?

Answer 58

Some series have a range of validity (a domain for x if you like).

eg ln(1+x) is valid for x between -1 and 1.

Answer 59

wait im going to answer that!

Answer 60

If you haven't done this, it won't be in your test...