Question

Taylor Polynomial Q

Answer 1

the form is a(x-c)^n right?

Answer 2

oh there's a -1 in there... a(-1)^n (x-c)^n?

Answer 3

and the function is defined as f(x)= a/x? centered at some c value...?

Answer 4

\[\Large \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n \]

Here centered at \(a\).

Answer 5

my book shows something different though, so I'm confused...

Answer 6

why centered @ a?

Answer 7

it shows I have to take the derivative like maclauren polynomials?

Answer 8

I'll just post this q.

Answer 9

\[\Large (x-a) \]
and \(a\) is a value where you want to approximate your function, so I consider that as it's center, the Maclaurin Series is always centered at zero.

Answer 10

Find the third taylor polynomial for f(x) = 7/x expanded about c= 1.

Answer 11

This is Maclaurin Series

\[\Large \sum_{n=1}^\infty \frac{f^{(n)}(0)}{n!} \]

Answer 12

the function and it's following derivatives evaluated at zero, rather then a.

Answer 13

all I have written down is a(-1)^n (x-c)^n

7(-1)^n (x-1)^n

7-7(x-1) + 7(x-1)^2 - 7(x-1)^3

Answer 14

hmm there q is prob here on open study, I'll go find it.

Answer 15

http://openstudy.com/users/konradzuse#/updates/500de97fe4b0ed432e105b6b

Answer 16

\[ \Large \sum_{n=1}^\infty \frac{f^{(n)}(0)}{n!}x^n\] *Correction*
Forgot about the exponential term.

Answer 17

huh I guess I skipped a bunch of stuff when writing this down...

Answer 18

yeah I was going to say what happened to (x-c)^n :P

Answer 19

oic what happened he created his own form.. wtf....

Answer 20

so basically whatever it's centered at is what f(a)=? so for example in this one centered @ 1 we did f(1)?

Answer 21

yes (-:

Answer 22

ok I want to try to solve this again;..

Answer 23

how come they went to f(1)^4? shouldn't we only go to 3?

Answer 24

\[\Large \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n = \Large \sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!}(x-1)^n\]

Answer 25

so we need to now find f(1) f'(1) f''(1) and f'''(1)?

Answer 26

f(n)(1)=(−7)(−2)(−3)(−4)...(−n)/1n+1=(−1)n∗7∗n! how did he figure this last part out?

Answer 27

well it means basically that you want to go to degree 3

Answer 28

but the first one is just a constant, so degree of zero.

Answer 29

Well you can see a pattern emerging there for the constants

\[ (-7)(-2)(-3)(-4)(-5) \cdot \dots \]
Factor out a minus one and you can see that the 7 stays fix, the rest is n factorial.

Answer 30

the term \[(-1)n\] Is just the algebra which says that you can factor out minus one n times.

Answer 31

I guess it's a lot of puzzle solving...

Answer 32

to create anew form...

Answer 33

or we could just use the old form and solve?

Answer 34

in almost all the taylor polynomial problems you have to find a pattern, which is indeed some kind of puzzle solving (-: It's easier for Maclaurin Series especially if you are allowed to use a cheat sheet.

Answer 35

I'm allowed to bring a sheet in yeah.

Answer 36

gimmeh teh cheat codes... down up, left right ab.

Answer 37

hehehe, for Taylor Polynomials they wont help you much, they are all computed for Mac Laurin, which as I said, are more straightforward.

Answer 38

okay so the formula means NOTHING then huh??????

Answer 39

as long as we have our f(a) and c = ?? right?

Answer 40

The formula is the definition of the Taylor Series, but in every case, you have to compute a few values (often more then 2, approximately 4+) so you can see a pattern and then you can find a general solution to it, which is what @mukushla did in the question you have linked above.

Answer 41

yessir, but he didnt' even use the formula. Once we knew it was centered @ 1 and the formula we just plugged it right in... 7/1 7/1^1 7/1^3 etc....

Answer 42

this poop asks for a polynomial f(x) = tan(x) c = pi/4.

so now we have f(pi/4) = tan(pi/4) = 1

Answer 43

Yes I understand what you mean, but he still followed the general pattern how you computer a Taylor Series, the center and the function itself just makes it easy in this case (dividing by one) but that clearly isn't always the case.

You always start the same way, for Taylor Series and MacLaurin Series, you evaluate the function at the given point and then you evaluate all it's derivatives at the given point.

Answer 44

f'() = 1+x?

Answer 45

f''(x) 1+x+x^2?

Answer 46

f'''() = 1+x+x^2+x^3?

Answer 47

hmph just read what you wrote,so you have to plug it into the formula to get the start?

Answer 48

The derivative of tan(x) is sec^2(x)

Answer 49

evaluated at pi/4 is equal to 2.

Answer 50

:O

Answer 51

alrighty, lol shouldn't have just relied on wolfram for that ;p.

Answer 52

so f(pi/4) = tan(pi/4) = 1

f'(pi/4) = sec^2(pi/4) = 2?

Answer 53

yes

Answer 54

f''(pi/4) = 2sec(pi/4) sec(pi/4) tan(pi/4)?

Answer 55

4?

Answer 56

\[\Large f''(a) =2\tan a\sec^2a\]
For a=pi/2 it equals four yes.

Answer 57

The derivatives of this polynomial get awkwardly complicated, but they seem to double .

Answer 58

where'ds you get 2tanasec^2(a)? Yeah it's ridiculous...

Answer 59

so when would we need the forumula, I'm not getting that part... It seems like we are just plug and chugging...

Answer 60

Well considering the formula we care to know what

\[ \large f^{(n)}(a)\] is equal to right?

Answer 61

mhm

Answer 62

At first I wasn't sure what that meant, but n is just the amount of derivation.

Answer 63

and if a is 1 then there is nothing to change.

Answer 64

yes (-:

Answer 65

and what about (x-a)^n and the n!?

Answer 66

Idk even though a isn't 1 now, we still got the same answer, even though yours was different than mine...

Answer 67

sooo konfused :(

Answer 68

well have you found a representative for \[ f^{(n)}(a)\] yet?

Answer 69

it has to match the following pattern (we evaluated some of these, so I will write them down abbreviated)

1
2
4
8
16
32

Answer 70

hmm oic we have to plug it back into the original :O.

Answer 71

of course (-:

Answer 72

well f(a)^0 = 1 f(a)^1 = 2, f(a)^2 = 4

Answer 73

Well that's what I was asking this entire time wtf we are doing LOL.....

Answer 74

It seems to me like this will work

\[\Large f^{(n)}\left(\frac{\pi}{4}\right)=2^n \]

Answer 75

yeah true...

Answer 76

I was thinking of some weird form :P

Answer 77

so now we take the oriignal form f^n(a) * 2^n/n! (x-c)^n?

Answer 78

nvm 2^n/n! (x-1)^n

Answer 79

By the way, make sure that you (we) understand the question correct, if they are not asking for a general answer, but only for an approximation (to a certain) degree, it is much easier when we just plug in and evaluate the problem.

Answer 80

"Find the Taylor Polynomial for"

Answer 81

this next one says "Find the third Taylor polynomial"

Answer 82

samething.

Answer 83

well, okay, but the both questions aren't the same thing, I am just mentioning this because sometimes it can be really really hard (for taylor polynomials) especial to find a general solution.

However, you can always compute the first few derivatives and plug in the values and then back substitute.

Answer 84

But sometimes Taylor Polynomials maybe have a pattern like this:

1
2
4
16
80
242

Hence they only get computed by computers.

Answer 85

mhmhm

Answer 86

the questions have always asked at what point, so it's not too terrible...

Answer 87

2^n/n! (x-1)^n is this the correct form?

Answer 88

I would say it is

\[\Large \sum_{n=0}^\infty \frac{2^n}{n!}\left(x-\frac{\pi}{4}\right)^n \]

Answer 89

yeah pi/4 sorry...

Answer 90

Yes, just keep in mind that I speculated about the pattern, because I was too lazy to compute all this terrible derivatives like

\[ \Large f^{''}=2\tan x \sec ^2 x \\ \Large f^{'''}=-((\cos(2x)-2)\sec^4(x)) \]

Answer 91

how did you get f'' I got 2sec(x) sec(x)tan(x)?

Answer 92

that's why I asked if the problem might just wants you to have a degree of three evaluation of the polynomial.

Answer 93

same thing

\[2 \sec(x) \cdot \sec(x)\tan(x) = 2 \sec^2(x)\tan(x) \]

Answer 94

oh.. LOL

Answer 95

derp....

Answer 96

but those don't really matter now that we got the formula right ?

Answer 97

\[\Large \sum_{n=0}^\infty \frac{2^n}{n!}\left(x-\frac{\pi}{4}\right)^n\]

Answer 98

how do we start this if 2^0/0! doesn't exist?