Estimate integral without using Taylors formula

[Mampenda]

We are examining how to estimate integrals we can't estimae using the usual rules (Taylor's formula).
Use the fact that 2 < e < 3 and that g(x) = e^x is increasing. (You do not have to explain this, it is given in the exercise)
A)
1. Explain that for all real numbers x and all natural numbers n
e^(e^x) = 1+ e^x + (e^(2x))/2 + (e^(3x))/5 + ... + e^(Sx) *((e^((n+1)x) )/(n+1)! for an S in (0, e^x)
2. Explain that
( e^(n+1) - 1 )/(n+1)(n+1)! < \integral_0^1{ (e^(Sx) * \frac{ e^((n+1)x){(n+1)!} < (3^3) * \frac{ e^(n+1) - 1 }{ (n+1)(n+1)! }

B)
Find n such that 27/2 * \frac{ e^(n+1) - 1 }{ (n+1)(n+1)! } < 0.05

C)
Use the information gained from (A) and (B) to calculate the integral_0^1 e^(e^x) dx with an error less than 0.05.

 

[Deleted member 4993]

We are examining how to estimate integrals we can't estimae using the usual rules (Taylor's formula).
Use the fact that 2 < e < 3 and that g(x) = e^x is increasing. (You do not have to explain this, it is given in the exercise)
A)
1. Explain that for all real numbers x and all natural numbers n
e^(e^x) = 1+ e^x + (e^(2x))/2 + (e^(3x))/5 + ... + e^(Sx) *((e^((n+1)x) )/(n+1)! for an S in (0, e^x)
2. Explain that
( e^(n+1) - 1 )/(n+1)(n+1)! < \integral_0^1{ (e^(Sx) * \frac{ e^((n+1)x){(n+1)!} < (3^3) * \frac{ e^(n+1) - 1 }{ (n+1)(n+1)! }

B)
Find n such that 27/2 * \frac{ e^(n+1) - 1 }{ (n+1)(n+1)! } < 0.05

C)
Use the information gained from (A) and (B) to calculate the integral_0^1 e^(e^x) dx with an error less than 0.05.

Please show us what you have tried and exactly where you are stuck.
Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this problem.

 

[Mampenda]

Please show us what you have tried and exactly where you are stuck.
Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this problem.
View attachment 23154

For A1, I subsituted e^(e^x) with u and showed that we get the same inequality when substitutig back.

For A2 I used the facts given, which means that g(x1)<g(x2) when x1<x2 for x1,x2€ I, and that g′(x)>0 for all x. Then I put x=1 such that ex€(2,3)=I. If we remove the coefficients e^(Sx) and 3^3 last two parts of the inequality are identical. Then, for n=e we get:

e^(e^e) + e^x + {e^(2x)}/2 + {e^(3x)}/6 +...+ {e^(ex))}/e! + {e^(Sx)} * {e^(n+1)x}/(n+1)! For an S€(0,ex)S€(0,ex)

And for (0<x<1) we have 1< e^(Sx) < e^x < e^(e^1) < 3^3, hence we see that the inequality holds.

I feel like I have answered this part of the assignment correct, but if I have made a mistake I would appreciate being corrected and shown the right way.

But my question is for part B and C: I am to find an n (I assume using brute force because we are trying to use an alternative method than using Taylor's formula) such that

(27/2)∗{e^(n+1)−1}{(n+1)(n+1)!} < 0.05

by using the info gathered from A1 and A2 to calculate the integral (C) with an error less than 0.05. My thought here is to bruteforce to find the n and then substitute e^(e^x) with what I found in A (i.e. the Taylor Polynomial) and then calculate the integral and find an error <0.05 belonging to n. But how do I do this in practice?