Answer:
[tex]\mathbf{1 + \dfrac{x^2}{2!}+ \dfrac{5}{24}x^4+\dfrac{61}{720}x^6}[/tex]
Step-by-step explanation:
From the given information:
we are to find the first four nonzero terms of the Maclaurin series of the given function.
Sec x.
If we recall ; we will realize that the derivative of sec x = [tex]\dfrac{1 }{cos \ x}[/tex]
Also; for cos x ; the first four terms of its Maclaurin Series can be expressed as ;
=[tex]1- \dfrac{x^2}{2!}+\dfrac{x^4}{4!}- \dfrac{x^6}{6!}+...[/tex]
However, using the long division method: we have;
[tex]\dfrac{1}{1- \dfrac{x^2}{2!}+\dfrac{x^4}{4!}- \dfrac{x^6}{6!}}[/tex]
the rule of the long division method is to first use the 1 from the denominator to divide the 1 from the numerator. the multiply it with the answer we get which is (1) before subtracting it from that answer (1).
i.e
1/1 = 1
1 × 1 = 1
1 - 1 = 0
Afterwards; we will subtract the remaining integers from this numerator.
So, we have:
[tex]\dfrac{-(1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}- \dfrac{x^6}{6!} )}{0+ \dfrac{x^2}{2!}-\dfrac{x^4}{4!}+ \dfrac{x^6}{6!}}[/tex]
We are going to apply the same process to the remainder [tex]\dfrac{x^2}{2!}[/tex];
which is to divide the second integer with 1
[tex]\dfrac{\dfrac{x^2}{2!}}{1}= \dfrac{x^2}{2!}[/tex]
Then we will multiply the numerator with [tex]\dfrac{x^2}{2!}[/tex] ; the result will then be subtracted from the polynomial.
[tex]= \dfrac{-( \dfrac{x^2}{2!} - \dfrac{x^4}{2! 2!} + \dfrac{x^6}{2! 4!}- \dfrac{x^8}{2! 6!}) }{0 + \dfrac{5}{24}x^4 - \dfrac{7}{360}x^6- \dfrac{x^8}{2!6!} }[/tex]
Repeating the same process for remainder [tex]\dfrac{5}{24}x^4[/tex]; we have:
[tex]\dfrac{ \dfrac{5}{24}x^4 }{1}= \dfrac{5}{24}x^4[/tex]
so; we will need to multiply 1 with [tex]\dfrac{5}{24}x^4[/tex] and subtract it from the rest of the polynomial
[tex]=\dfrac{{0 + \dfrac{5}{24}x^4 - \dfrac{7}{360}x^6- \dfrac{x^8}{2!6!} }}{ 1-x^2+ \dfrac{x^4}{4!} - \dfrac{x^6}{6!} }[/tex]
[tex]= \dfrac {- ( \dfrac{5}{24}x^4 -\dfrac{5}{2!4!}x^6 - \dfrac{5x^8}{4!4!} - \dfrac{5x^{10}}{6!4!} } {0+ \dfrac{61}{720}x^6}[/tex]
Here ; the final remainder is [tex]\dfrac{61}{720}x^6}[/tex]; repeating the usual process for long division method; we have:
[tex]\dfrac{\dfrac{61}{720}x^6}{1}= \dfrac{61}{720}x^6}[/tex]
So;
[tex]= \dfrac{0+ \dfrac{61}{720}x^6}{1-x^2+ \dfrac{x^4}{4!} - \dfrac{x^6}{6!} }[/tex]
[tex]= \dfrac{-( \dfrac{61}{720}x^6)}{0 }[/tex]
Now the first four nonzero terms of the Maclaurin series is the addition of all the integers used as remainders ; i.e
[tex]\mathbf{1 + \dfrac{x^2}{2!}+ \dfrac{5}{24}x^4+\dfrac{61}{720}x^6}[/tex]
