Question

Find the Taylor polynomials of degree n approximating

((4)/(5-5x))

for x near 0

For n=3, P3(X)=
For n=5, P5(X)=
For n=7, P7(X)=

@agent0smith

Answer 1

i would write it as
\[\frac{4}{5}\frac{1}{1-x}\] and use the expansion of the geometric series ,
unless i am reading it wrong

Answer 2

how would you do that?

Answer 3

\[\frac{1}{1-x}=1+x+x^2+x^3+...=\sum_{k=0}^{\infty}x^k\]

Answer 4

a very frequent gimmick rather than taking successive derivatives which is usually a pain, is to arrange the thing you have to look like a geometric series somehow, or the derivative of one

Answer 5

so then what do i do with that series?

Answer 6

multiply it by \(\frac{4}{5}\) then stop for whatever degree polynomial you want

Answer 7

so for example \(P_3\) would be
\[\frac{4}{5}(1+x+x^2+x^3)\]

Answer 8

and thats the answer?

Answer 9

i think so, yes

Answer 10

yay it was right:)

Answer 11

it is like all those problems you did where you had to sum a geometric series, only this time you are working backwards
you have the closed form you write it in the open form

Answer 12

whew, i am glad i didn't screw that one up

Answer 13

im definitely going to need some practice with these lol :/

Answer 14

thank you!

Answer 15

yw

Answer 16

how would i do it with sin(3x)?

Answer 17

memorize the power series for \(\sin(x)\) then replace \(x\) by \(3x\)

Answer 18

do you know the power series for \(\sin(x)\) ?

Answer 19

no i don't:/

Answer 20

ooooh that is a problem
ok it is easy to remember
first of all sine is odd, so the power series has only odd exponents

Answer 21

and it alternates
and the denominators are \(n!\) just like for \(e^x\)

Answer 22

okay:)

Answer 23

\[x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\]

Answer 24

oh wait doesnt it just alternate between 0,1,0,-1.....

Answer 25

when i said alternate, i was ignoring the terms that are not there

Answer 26

but yes, you can think of it that way if you like

Answer 27

i like urs better.....so then how do i do it when its 3x not just x? do i just add 3x to the numerator?

Answer 28

oh i see what you're saying
the derivatives evaluated as zero alternate between those numbers
so yes, you know it and therefore you know the power series as well

Answer 29

oh no i dont know anything with this chapter lol it just sounded familiar! this is the worst chapter my professor has taught so far!

Answer 30

\[sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\]
\[\sin(3x)=3x-\frac{(3x)^3}{3!}+\frac{(3x)^5}{5!}-\frac{(3x)^7}{7!}+...\]

Answer 31

yes that is exactly what i meant lol!

Answer 32

yeah, then you are right
i have to say this is tons easier than some of the stuff you have been doing

Answer 33

so its P1=3x?

Answer 34

yeah

Answer 35

P3=3x-(((3x)^3)/(3!))+(((3x)^5)/(5!))

Answer 36

P5=3x-(((3x)^3)/(3!))+(((3x)^5)/(5!))-(((3x)^7)/(7!))+(((3x)^9)/(9!))

Answer 37

i thought P3 meant the polynomial of degree 3

Answer 38

not the one with three terms

Answer 39

oh shoot ur right....
so then how would i do it?

Answer 40

degree 3 one is
\[3x-\frac{(3x)^3}{3!}\]

Answer 41

and degree 5 one is the one you wrote as the degree 3 one

Answer 42

how did u know that?

Answer 43

jk i see it now

Answer 44

because
\[3x-\frac{(3x)^3}{3!}\] is a polynomial...
ok here i was going to explain it

Answer 45

thank you very much you were a lot of help:) so ahead and explain though!

Answer 46

you might want to reduce the fraction and write is as
\[3x-\frac{3x^3}{2}\]

Answer 47

okay:)

Answer 48

you are welcome, and no i will not explain why
\[3x-\frac{3x^3}{2}\] is a polynomial of degree 3 !
good luck with this, it is not that bad