Answer :
The number of terms of the Maclaurin series for eˣ that should be used to estimate [tex]e^{0.1}[/tex] to within 0.00001 is 4.
Taylor’s inequality:
If Rn( x ) = 0
R_n(x)=0
Rn(x)=0.
The given function is eˣ
Thus, all the derivatives of eˣ is eˣ itself.
Suppose that, the nth derivative of eˣ is eˣ
To estimate the value of [tex]e^{0.1}[/tex] substitute the value of x as 0.1.
Note that Taylor series is called Maclaurin series when a=0.
Usually, Maclaurin series will starts with n=2.
Increase the value of n until the reminder is less than 0.00001.
By Taylor’s inequality,
If | [tex]f^{n}[/tex](0.1) | = [tex]e^{0.1}[/tex]≤ M then | Rₙ(x) |≤[tex]\frac{M}{(n+1)!}[/tex] [tex]|0.1-0|^{n+1}[/tex]
Substitute the value of n as 2. Then,
[tex]|R2(x)|[/tex]≤[tex]\frac{M}{(2+1)!} |0.1-0|^{2+1}[/tex]
≤ [tex]\frac{e^{0.1}}{3!} |0.1|^{3}[/tex]
≤ [tex]\frac{e^{0.1}}{6} 0.001[/tex]
≤ 0.00018
Since [tex]|R2(x)|[/tex] ≥ choose the value of n as 3. Then,
[tex]|R3(x)|[/tex] ≤[tex]\frac{M}{(3+1)!} |0.1-0|^{3+1}[/tex]
≤ [tex]\frac{e^{0.1}}{4!} |0.1|^{4}[/tex]
≤ [tex]\frac{e^{0.1}}{24} 0.0001[/tex]
≤ 0.0000046
Therefore, [tex]|R3(x)|[/tex] ≤ 0.00001
That is, n=3 satisfies the inequality.
So, by adding the 4 terms (when n=0,1,2,and 3) of the Maclaurin series, it
is possible to estimate the value of [tex]e^{0.1}[/tex] to within 0.00001.
Thus, the number of terms of the Maclaurin series for eˣ that should be used to estimate [tex]e^{0.1}[/tex] to within 0.00001 is 4.
The number of terms of the Maclaurin series for eˣ that should be used to estimate [tex]e^{0.1}[/tex] to within 0.00001 is 4.
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