Question

Determine the degree of the Maclaurin polynomial required for the error in the approximation of e to be less than 0.00005.

Answer 1

a) 1
b) 3
c) 5
d) 7

I chose D becuase of the Formula \[\Large \left| S_k-S \right|\le a_{k+1} \le 5 \times 10^{-5}\]

Answer 2

Sorry Wrong formula, thats for an alternating series.
\[\Large \left| E_n(x) \right| \le \frac{M}{(n+1)!}\left| x-a \right|^{n+1}\]

Answer 3

Where \[\Large \left| E_n(x) \right|= 5 \times 10^{-4}\]
\[\Large M=\text{Maximum value of }~\left| f^{n+1}(x) \right|\]

Answer 4

\[\Large \left| E_n(x) \right| \le \frac{e}{(n+1)!} \le 5 \times 10^-5\]
n=7
can someone verify this?

Answer 5

x=1 since we're evaluating e^(1) and the a=0 since this maclaurin and that is just 1 to a power and it just goes away, I know you probably know that, but I didn't just gget rid of it

Answer 6

The remainder term, Rn(x), for a Taylor's expansion about a equals
\(R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\)
where c is a number between a and x.

Here
a=0 (McLaurin's series)
x=1 (e^1)
\(0\le c\le1\)
\(f(x)=e^x\)
\(f^{(n)}(x)=e^x\) for all n
Since \(f^{(n)}(x)=e^x\) is a strictly increasing function, we evaluate Rn(x) at x=1, which simplifies Rn(x) to
\(R_n(1)=\frac{e}{(n+1)!}\)
However, unfortunately
\(R_5(1)=0.003775\)
\(R_7(1)=6.74\times 10^{-5} > 5\times 10^{-5}\)

Answer 7

Yeah \(T_7(x)\) is keeps the error under \(5\times 10^{-4}\) only according to remainder thm, perhaps there is a typo in the main question

Answer 8

Ok, so what would be the answer? Becaus 7 is the highest answer given and if it's not the answers larger than than that

Answer 9

Honestly, I'll just go with 7

Answer 10

7 is correct

Answer 11

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Answer 12

i thought the question is about remainder thm

Answer 13

How come wehn we use the formula then, it isn't <5 x 10^(-5)

Answer 14

@mathmate 's work shows why the 7th degree fails the remainder thm

Answer 15

What do you mean fails? And also, then why is it that the exp-actual is < 5 x 10^-5 when you did it

Answer 16

you can use the remainder theorem to bound the error

Answer 17

\(R_n(1) = \dfrac{e}{(n+1)!}\)

you want to pick "n" value such that this remainder is less than the desired error

Answer 18

remainder theorem gives you a bound, that bound need not be the optimal error

Answer 19

Remainder theorem tells you that your height is with in 1 feet of 6,
your actual height could be 6.2

Answer 20

example^

Answer 21

Oh, so it says it is within this range, but it doesn't mean that it isn't less, as seen in the example of @perl

Answer 22

What i posted was the exact error. the remainder theorem gives you a bound on the exact error, so you can think of it as an approximation of the error.

Answer 23

So 7 terms, as seen in his example, has an eror that is less than 0.00005, but the bound (so it can't be greater than) is 6.7 x10^(-5)

Answer 24

perl just worked it bruteforce (he calculated ur height using a measuring tape)

but you want to use remainder theorem which gives you info like : your height will be within 1 foot of 6

Answer 25

Thanks

Answer 26

It looks like the remainder theorem could not be used in this problem, which is strange.

Answer 27

btw, 7 is wrong according to remainder thm

Answer 28

why remainder thm cannot be used ?

Answer 29

another way to describe the remainder theorem is that it gives you a tolerance on the exact error

Answer 30

there might be a typo

Answer 31

or the person who made the question did not anticipate that the remainder theorem would not work (did not actually work it out). but it is true that 7th degree taylor will work

Answer 32

its possible that the person who made the question did what i just did :)

Answer 33

or some calculator

Answer 34

remainder thm says that the 8th degree polynomial works, which looks fine to me

Answer 35

Yeah, the people who make my lessons and the people who make the tests aren't the same company sadly, so I've been doing remainder theorem while my lesson calls is the lgrange error bound and I finally undersand what these questions are saying xD

Answer 36

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Answer 37

\(R_8(1) = \dfrac{e}{(8+1)!}\lt 7.5\times10^{-6}\lt 5\times 10^{-5} \)

Answer 38

I agree with the "speculation" that the person making up the equation calculated using the exact error using "brute force". After all, the question did not mention using the remainder theorem.

This agrees very well with the fact that the lesson and exercise are not coordinated.

In this case, we all know how to calculate the exact error, which is why it would have been an excellent example to verify and understand the remainder theorem. However, as it is, it does illustrate the important conceptual difference between the upper bound and the exact error.