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Use multiplication or division of power series to determine the first three non-zero terms of the Maclaurin Series for the following functions. 50. f(x)=sec(x) 51. f(x)=e x

ln(1−x)

User Mistika
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Sure, I'd be happy to walk you through the process of finding the first three non-zero terms of the Maclaurin Series for these functions.

50. f(x)=sec(x)

To find the Maclaurin series of sec(x), we recognize that the secant function is the reciprocal of the cosine function. The Maclaurin series for cos(x) is
1 - x^2/2 + x^4/24 - x^6/720 + ...

Taking the reciprocal of this results in the Maclaurin series for sec(x), but as we are interested only in the first three non-zero terms (i.e up-to x^2), we will get term for x^0 and x^2.

So therefore, the Maclaurin series up to x^2 for sec(x) is 1 + x^2/2

51. f(x)=e^x * ln(1-x)

For this problem, we need to cross-multiply the Maclaurin series expansions of e^x and ln(1 - x).

The Maclaurin series for e^x is
1 + x + x^2/2 + x^3 / 6 + ...

And the Maclaurin series for ln(1 - x) is
-x - x^2 / 2 - x^3 / 3 - ...

With simple multiplication and reduction, we can find the terms for the combined series.
For the first term, no multiplication is needed as base term of e^x (1) multiplies with the base term of ln(1 - x)(-x) which results in -x.
Second term is found by:
(second term of e^x which is x) * (base term of ln(1 - x) which is -x) + (base term of e^x which is 1) * (second term of ln(1 - x) which is -x^2 / 2)
This simplifies to -x^2 - x^2 / 2 which results in -3x^2 / 2.
We stop here because the question asks only for the first three non-zero terms.

So, the Maclaurin series up to x ^ 2 for e^x * ln(1 - x) is -x - 3x^2 / 2.

User Daniel Higueras
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