Lagrange Remainder question will be posted below! Thank you!

- mortonsalt

Lagrange Remainder question will be posted below! Thank you!

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- mortonsalt

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:
\[|R_n| \le\]

- amistre64

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

- amistre64

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

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## More answers

- mortonsalt

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

- amistre64

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

- amistre64

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

- mortonsalt

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

- mortonsalt

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

- amistre64

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

- amistre64

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

- mortonsalt

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 Taylorâ€™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

- amistre64

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

- mortonsalt

T_1 = 10/3
T_2 = 179/54
I'm also not entirely sure why it's an "underestimate" of f(10).

- mathmate

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

- mathmate

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.

- mathmate

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

- mathmate

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,
\(R_2 = \frac{f^{(3)}(c)}{(3)!} (10-8)^{(3)}\)
\(R_2 = \frac{f^{(3)}(c)}{(3)!} (2)^{(3)}\)

- mortonsalt

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

- mathmate

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.

- mathmate

|dw:1446868262880:dw|
Hope this explains how we find c in this particular case.

- mathmate

Also, recall that R2 is an upperbound estimate.

- mathmate

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

- mortonsalt

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

- mathmate

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.

- mortonsalt

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

- mathmate

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

- mathmate

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.

- mathmate

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

- mortonsalt

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?

- mathmate

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

- mathmate

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

- mathmate

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

- mortonsalt

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?

- mortonsalt

x not c, oh god im losing it

- mathmate

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.

- mortonsalt

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

- mathmate

No problem! :)

- mathmate

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

- mathmate

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
\(R_2 = (f'''(8)/3!)(10-2)^3\)

- mortonsalt

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

- mathmate

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

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