Question

I just need someone to tell me what this question wants in dummy terms: Suppose that p is a polynomial of degree d. Given any \(x_0\in R\), show that the \((d+1){th}\) Taylor Polynomial for p at \(x_0\) is equal to p.

Answer 1

I don't understand how you can have a (d+1)th for a d degree function.

Answer 2

@SithsAndGiggles

Answer 3

d+1 degree Taylor polynomial, do you know what a taylor polynomial is?

Answer 4

i think you need to show that after differentiating a polynomial d+1 times, you get a zero term and all remaining terms will be 0

Answer 5

yes, it's more of a the d+1 will end with a 0. because past d the deriv is zero. So I don't understand what it's asking

Answer 6

oh, thanks ganeshie. I guess my reason for confusion is what I am supposed to show

Answer 7

what it is saying if we plug in x0 for x in the polynomial of degree d it should equal the same thing as a taylor series of degree d+1

Answer 8

yeah i have no clue yet how to start the proof too, but it is clear what we need to prove

Answer 9

I know how to prove it, but I just didn't understand the problem thanks

Answer 10

can you medal Miracrown for helping for me ganeshie?

Answer 11

no medal me

Answer 12

already gave it to mira :)
fib you medal dan maybe..

Answer 13

<3

Answer 14

and x0 is a real number

Answer 15

i medal ganeshi

Answer 16

ah basically its a matter of taylor series expansion being the approimations wrt to derivatives

Answer 17

for a certain nth degree polynomial all the succeding taylor approximations after d+1 yeild zero so

Answer 18

how to start a proof like this

Answer 19

it should be sufficient to just state it in words right ?

Answer 20

i am going with based off of the definition of the taylor p, if a deriv =0 then the term of the poly=0 so if we show d+1,2,3 by induction =0 then we are left with the taylor of d

Answer 21

all polynomials can be rewritten in form
kth polynomial = ax^k+bx^(k-1)....
Therefore
the k+1 derivative of this kth polynomial = 0

Answer 22

(k+1)th derivative

Answer 23

consider a polynomial of degree "d"
by taylor theorem, the taylor series equals this polynomial. however after differentating this polynomial for d+1 times, we get f^(d+1) = 0 and thus all the subsequent will be 0. that means the first d+1 terms of taylor series equals the polynomial \(\square\)

Answer 24

y'all want some more? I have like 4 more due tomorrow at 1 that I need interpretations so I understand what it is asking

Answer 25

ya only if u gimme ur medals this time

Answer 26

but you don't have to complete them like you are

Answer 27

bahahahah

Answer 28

as told by ganeshie, I have passed the medal

Answer 29

Also, I hate analysis. Any help is appreciated

Answer 30

just a quick q, does taylor series exists for all functions ?

Answer 31

well, essentially it is the MVT for derivatives, so so long as it's differentiable, I think so

Answer 32

consider a polynomial of degree "d"
by taylor theorem, the taylor series, `if it exists`, equals this polynomial. however after differentating this polynomial for d+1 times, we get f^(d+1) = 0 and thus all the subsequent will be 0. that means the first d+1 terms of taylor series equals the polynomial

Answer 33

yeah its redundant if we agree that the series exists for all polynomials

Answer 34

as all polynomials are differentiable

Answer 35

def exists for all poly, but what about trancendentals? (sp? sorry)

Answer 36

what does a transcendental mean

Answer 37

A transcendental function is a function that does not satisfy a polynomial equation whose coefficients are themselves roots of polynomials, in contrast to an algebraic function, which does satisfy such an equation.

Answer 38

hey it's essentially p=np.... freaky

Answer 39

dp/dn= p+dp/dn
D_2 = dp/dp + d^2p/dn^2
D_N= d^(n-1)/dp^(n-1) +d^np/dn^n

Answer 40

all derivatives exist

Answer 41

so ud need infinite taylor approximations

Answer 42

yep like sin and cos

Answer 43

yah

Answer 44

sin and cos are infinite polynomials kinda

Answer 45

like u know ax^4+bx^3... looks like this w