Question

so what does the "a" contribute in a taylor series?

P(x) = f(a)+f'(a)(x-a)+f''(a)(x-a)^2 ...

Answer 1

its the x0 ... when they say find the taylor polynomial about a point x0

Answer 2

right .... they call it a basis point in some texts; and when it modifies the x part its translates into a shift in the graph ... I was curled up with cozy calculus book last night and finally realized what a taylor series was. Before that it was rather enigmatic to me. But I was curious about the x-a parts ...

Answer 3

can i ask yu a question now?...(A)

Answer 4

I made a cos(x) polynomial with a=0 .. but was curious about what would happen if I made a= something else but feel asleep and dreamed of rabid gators or some such :)

Answer 5

you can ask

Answer 6

are you good at numerical analysis?

Answer 7

i dont know; i might be, but I never sat down and tried to determine what numerical analysis actually is :)

Answer 8

and btw when a=0 its calld a maclaurin series

Answer 9

ok i'll show u the question,

Answer 10

nice :) I found out last night that the poly expressed the degrees in radians tho ... would our a=the radian version of 42 then?

Answer 11

yess which is 7pi/30

Answer 12

and would we be polying up this with the derivatives of cos(pi/4) ??

Answer 13

sorry to be rude ha but amistre i still need help man could you help when u done here

Answer 14

cos(a) = 1/sqrt(2)
-sin(a) = -1/sqrt(2)
-cos(a) = 1/sqrt(2) ... like that?

Answer 15

ill take a look at it when i can :)

Answer 16

im confused :( :'(

Answer 17

cos(d-a) then.... that doesnt seem like an easy thing to check unless you already know the sin and cos amounts for (7pi/30-pi/4) ...

Answer 18

28/120
30/120 are pretty close, so id assume its evalulated at pi/4

Answer 19

i hav the solution but its confusing me, im stuck

Answer 20

lets use a=pi/4 (30pi/120) to approximate a poly for cos(7pi/30), (28pi/120) ok?

Answer 21

okaaay

Answer 22

lets take a generic poly set up and derive it down then we can equate it to the derivaites of cos(45):

\[P(x) = ax^8+bx^7+cx^6+dx^5+ex^4+fx^3+gx^2+hx+i\]
\[P1(x) = 8ax^7+7bx^6+6cx^5+5dx^4+4ex^3+3fx^2+2gx+h\]
\[P_1(x) = 8ax^7+7bx^6+6cx^5+5dx^4+4ex^3+3fx^2+2gx+h\]
\[P_2(x) = 56ax^6+42bx^5+30cx^4+20dx^3+12ex^2+6fx+2g\]
\[P_3(x) = 336ax^5+210bx^4+120cx^3+60dx^2+24ex+6fx\]
etc right?

Answer 23

eventually we get down to factorials; since each coefficient is the product of each successive deriving: 8.7.6.5.4.3.2.1 for the x^8 part right?

Answer 24

thats what i got
first i find the exact value of cos 42 7 digits
then i use taylor poly till thers a difference a number in 10^-6
then cos 42 - P3(x) = error
and thats it :S

Answer 25

0. 1/sqrt(2)
1. -1/sqrt(2)
2. -1/sqrt(2)
3. 1/sqrt(2)
4. 1/sqrt(2)
5. -1/sqrt(2)
6. -1/sqrt(2)
7. 1/sqrt(2)
8. 1/sqrt(2)

\[ [cos(45)]_8=P_8=8!a=1/\sqrt{2};\ a=1/8!\sqrt{2}\]
\[ [cos(45)]_7=P_7=\frac{8!}{8!/\sqrt{2}}+7!b=1/\sqrt{2}\]
\(=\sqrt{2}+7!b=\cfrac{1}{\sqrt{2}}\)
\( 7!b=\cfrac{1}{\sqrt{2}}-\sqrt{2}\)
\( b=\left(\cfrac{1}{\sqrt{2}}-\sqrt{2}\right)/7! \)
that gets messy real quick like it seems :)

Answer 26

and typing it in latex format is a pain...

Answer 27

at any rate; that is where we get the factorial denominators from

P(x)=\(\frac{}{0!}+\frac{}{1!}x+\frac{}{2!}x^2+\frac{}{3!}x^3+\frac{}{4!}x^4+\frac{}{5!}x^5+\frac{}{6!}x^6+\frac{}{7!}x^7+\frac{}{8!}x^8\)

just curious; but is this when we use the (x-a)^# stuff?

Answer 28

yesss

Answer 29

and the tops of the fractions here equal the derivative values right?

f(a), f'(a), f''(a), ... correct?

Answer 30

yess correct

Answer 31

\[P(x)=cos(a)-\frac{sin(a)}{1!}(x-\frac{\pi}{4})-\frac{cos(a)}{2!}(x-\frac{\pi}{4})^2+\frac{sin(a)}{3!}(x-\frac{\pi}{4})^3...\]

Answer 32

\[P(x)=\frac{1}{\sqrt{2}}-\frac{(x-\frac{\pi}{4})}{\sqrt{2}}-\frac{(x-\frac{\pi}{4})^2}{2!\sqrt{2}}+\frac{(x-\frac{\pi}{4})^3}{3!\sqrt{2}}+\frac{(x-\frac{\pi}{4})^4}{4!\sqrt{2}}-\frac{(x-\frac{\pi}{4})^5}{5!\sqrt{2}}\]

Answer 33

add more poly terms as needed :)

Answer 34

In a power series we talk about a power series being "centered" at a given point.
In many cases the power series won't converge absolutely on the entire domain, it will only converge within a "radius of convergence" and it does so centered at "a".

Answer 35

id go to 8 and test it out; but that many just goes off the screen

Answer 36

http://en.wikipedia.org/wiki/Radius_of_convergence

Answer 37

or stop at 5 and test it out :)

Answer 38

so the "a" is a way to move our poly to the desired position and fit it to our curve right?

Answer 39

Except in some cases the radius of convergence will be infinite and so it doesnt matter what "a" you choose.

Answer 40

im gonna have to plug this into me calculator to assess its veracity :)

Answer 41

the function will be "centered" around a im just saying it will converge everywhere if the radius of convergence is infinite.

Answer 42

you would need to graph it to check this.

Answer 43

my graphs dont line up when I do the x-a parts .....

Answer 44

try it for sin(x) then choose any a you like

Answer 45

\[\sum _{n=0}^{\infty }{\frac { sin^{ \left( n \right) }
\left( 10 \right) \left( x-10 \right) ^{n}}{n!}} = sin(x)\]

Answer 46

when I use the cos(0) for my derivative bits I get it spot on to 6 decimals

Answer 47

I just chose a = 10 but it doesn't matter

Answer 48

the - a in the factor (x-a) just offsets the "centering" in the choice of coefficients.
without it, it would be sin(x + a)

Answer 49

Anyways the trig functions sin, and cos have infinite radiuses of convergence.

Answer 50

Thats why "a" for those functions don't matter

Answer 51

ohhh.....