Respuesta :
Answer:
The estimate of In(1.4) is the first five non-zero terms.
Step-by-step explanation:
From the given information:
We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)
So, by the application of Maclurin Series which can be expressed as:
[tex]f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)[/tex]
Let examine f(x) = In(1+x), then find its derivatives;
f(x) = In(1+x)
[tex]f'(x) = \dfrac{1}{1+x}[/tex]
f'(0) [tex]= \dfrac{1}{1+0}=1[/tex]
f ' ' (x) [tex]= \dfrac{1}{(1+x)^2}[/tex]
f ' ' (x) [tex]= \dfrac{1}{(1+0)^2}=-1[/tex]
f ' ' '(x) [tex]= \dfrac{2}{(1+x)^3}[/tex]
f ' ' '(x) [tex]= \dfrac{2}{(1+0)^3} = 2[/tex]
f ' ' ' '(x) [tex]= \dfrac{6}{(1+x)^4}[/tex]
f ' ' ' '(x) [tex]= \dfrac{6}{(1+0)^4}=-6[/tex]
f ' ' ' ' ' (x) [tex]= \dfrac{24}{(1+x)^5} = 24[/tex]
f ' ' ' ' ' (x) [tex]= \dfrac{24}{(1+0)^5} = 24[/tex]
Now, the next process is to substitute the above values back into equation (1)
[tex]f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...[/tex]
[tex]In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...[/tex]
[tex]In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...[/tex]
To estimate the value of In(1.4), let's replace x with 0.4
[tex]In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...[/tex]
[tex]In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...[/tex]
Therefore, from the above calculations, we will realize that the value of [tex]\dfrac{0.4^5}{5}= 0.002048[/tex] as well as [tex]\dfrac{0.4^6}{6}= 0.00068267[/tex] which are less than 0.001
Hence, the estimate of In(1.4) to the term is [tex]\dfrac{0.4^5}{5}[/tex] is said to be enough to justify our claim.
∴
The estimate of In(1.4) is the first five non-zero terms.
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