task: In this task we will investigate how we can estimate integras we don't have a chance at by normalruels by using taylors formula. You can use 2 < e < 3 without justification and that the functiongiven g(x) = e* is growing.a) Justify that all real numbers x and positive integers n, so ise 2x= 1+ e* +e 3xenx+п!eet(n+1)xee Sx(n + 1)!For a sy E (0, ex). justify that2en+1_ 1(n+1)(n+1)!b) Find n so thate Sxe(n+1)x-dx < 33(n+1)!en+1–1(n+1)(n+1)!27en+112 (n + 1)(n + 1)!< 0.05?1Use the information above to calculateeetdx with an error less than 0.05

Answered: Justify that all real numbers x and…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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“Since you have asked multiple questions, we will solve the first question for you i.e. (a). If you want any specific question to be solved then please specify the question number or post only that question.”

(a)

We know that,

$e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . + \frac{x^{n}}{n!} + R_{n} (x) where, R_{n} (x) = \frac{e^{c}}{(n + 1)!} x^{n + 1} for c \in (0, x) .$

Step 2

We replace $x$ with $e^{x}$ and $c with e^{c},$ we get

$e^{e^{x}} = 1 + e^{x} + \frac{{(e^{x})}^{2}}{2!} + \frac{{(e^{x})}^{3}}{3!} + . . . + \frac{{(e^{x})}^{n}}{n!} + R_{n} (e^{x}) where, R_{n} (e^{x}) = \frac{e^{e^{c}}}{(n + 1)!} e^{x (n + 1)} for e^{c} \in (0, e^{x}) . In our case, S_{x} = e^{c} . or we can also say S_{x} \in (0, e^{x}) . ∴ e^{e^{x}} = 1 + e^{x} + \frac{e^{2 x}}{2!} + \frac{e^{3 x}}{3!} + . . . + \frac{e^{n x}}{n!} + R_{n} where, R_{n} = e^{S_{x}} \frac{e^{x (n + 1)}}{(n + 1)!}, S_{x} \in (0, e^{x}) .$

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