Question

Lagrange Remainder question will be posted below! Thank you!

Answer 1

Consider
\[f(x) = \sqrt{x+1}\]

Let \[T_n\] be the n^th degree Taylor Polynomial of f(10) about x=8.

If R_2 is the remainder given by the Lagrange Remainder Formula:
\[|R_n| \le\]

Answer 2

wouldnt we just take the integral from x=8 to inf? or am i thinking of some other remainder thrm stuff

Answer 3

something about the remainder is no greater then the n+1th term ?

Answer 4

This is under differential calculus, so I'm not entirely sure if that's what I'm supposed to do. :(

Answer 5

what does your material suggest? the google says that the terminology can be used to mean different things .... would hate to be discussing apples instead of oranges

Answer 6

http://www.stewartcalculus.com/data/CALCULUS%20Early%20Transcendentals/upfiles/Formulas4RemainderTaylorSeries5ET.pdf

Answer 7

The original question asked me for T_1 and T_2, and whether or not it is an under/over estimate. T_2 = 179/54 at f(10) about x=8. So I'm guessing they require me to use that information

Also \[R_n = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{(n+1)}\]

Answer 8

But I'm not entirely sure how to apply it.

Answer 9

\[\Large S=\int_{a}^{b}f(t)dt\]
\[\Large S=\underbrace{\int_{a}^{c}f(t)dt}_{S_n}+\underbrace{\int_{c}^{b}f(t)dt}_{R_n}\]

Answer 10

T_i being the taylor poly consisting of the first i terms?

Answer 11

Suppose that f has derivatives of at least order n + 1 on some interval [b, d]. Then if x and a are any two numbers in (b, d), the remainder Rn(x) in Taylor’s formula can be written like the formula I wrote above, where c is some number between x and a.

T_1 being the first Taylor Polynomial when f(10) is about x=8. @amistre64

Answer 12

x-a = 0, when x 'is about' a. i beleive. or rather is centered about a

Answer 13

T_1 = 10/3
T_2 = 179/54

I'm also not entirely sure why it's an "underestimate" of f(10).

Answer 14

Is it possible to post an image of the question?
In addition, the inequality \(|R_n|\le\) is not complete above.

Answer 15

Your notation of T1 is actually the sum of the first term and the constant term, and T2 is the sum up to the second term.
The first few terms are:
f(x)=sqrt(1+x)=3+(1/6)(x-8)+(-1/216)(x-8)^2
The sums are
S1=3+(1/6)(10-8)=10/3=3.333..... > 3.1622777...
S2=3+(1/6)(10-8)+(-1/216)(x-8)^2=179/54=3.3148148...<3.162777
S2 is an underestimate because the series is alternating in signs, so the sums would alternate around the correct value.

Answer 16

*correction 3.162777 should read 3.31662479, = sqrt(1+x)=sqrt(1+10)=sqrt(11)

Answer 17

Now you can evaluate the value of R2.
In your expression for Rn,
\(R_n = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{(n+1)}\)
c stands for a value between a and x such that abs(f^(n+1)(c)) is the largest.
Unfortunately, this is an upper bound and so way overestimates the error most of the time.
Substituting values,
\(R_2 = \frac{f^{(3)}(c)}{(3)!} (10-8)^{(3)}\)
\(R_2 = \frac{f^{(3)}(c)}{(3)!} (2)^{(3)}\)

Answer 18

@mathmate FINALLY ITS ALIVE, haha!
I'm still confused as to what "c" is exactly.

Answer 19

It turns out that f'''(x) is a decreasing function between 8 and 10, so the maximum value is at c=x0=8.
If you evaluate R2 with c=8, you will get R2=0.0020576
The actual error is, sqrt(11)-179.54=0.001809975
which shows that R2 is a rather good estimate in this case.

Answer 20

Hope this explains how we find c in this particular case.

Answer 21

Also, recall that R2 is an upperbound estimate.

Answer 22

Btw,
\(f^{(3)}(x)=f'''(x)=\frac{3}{8(x+1)^{5/2}}\)

Answer 23

@mathmate How do you know that it's a decreasing function?

Answer 24

You plot it and see. For a small interval, you could take a chance and calculate the two ends. But you can use a graphics calculator, or some online tools to plot it.

Answer 25

The thing is this course doesn't allow me to use a calculator at all! Which makes this problem ridiculously difficult to solve.

Answer 26

In this case, we also know that it is a decreasing function because the denominator is clearly an increasing function.

Answer 27

Yes, I noticed that the second term you gave 179/54, which is a good sign of hand calculations... kind of a rare gem these days.

Answer 28

\(f'''(8)=1/648\)

Answer 29

Does that mean that I have to get R1 and R2, check which one would have lesser/greater error, then choose the range of R_n according to this?

Answer 30

and \(R_2=(1/(648))/3! *(10-2)^3\)

Answer 31

No you only have to calculate R2, because 10 is in the radius of convergence, therefore more terms, more accurate.

Answer 32

\(R_2=(1/(648))/3! *(10-2)^3=1/486\)

Answer 33

Oh, sorry, that's not what I mean to ask, hahaha! I mean, calculate the R_2 when c=10 then when c=8? Or do we always have to choose the upperbound?

Answer 34

x not c, oh god im losing it

Answer 35

Yes, the remainder term is an upperbound estimate, so you have to choose the value of x between x0=8 and x=10 such that the absolute value of f^n(c) is the largest.

Answer 36

Oh god, I understand now. Thank you, thank you, thank you for your help and patience! I really appreciate it.

Answer 37

No problem! :)

Answer 38

* ...the value of \(\color{red}{c}\) between x0=8 and x=10...

Answer 39

Btw, just to make sure:
c applies ONLY to f^n(c) (c=8 for R2). The rest is x (=10 for this question)
so the complete expression for R2 is
\(R_2 = (f'''(8)/3!)(10-2)^3\)

Answer 40

@mathmate Alrighty! I assume you mean (10-8) instead of (10-2), btw. Thanks a bunch!

Answer 41

You're welcome! :)
yes, (10-8) is correct, finger faster than the brain, typed the result instead of the data! lol