Q. Write the first four non-zero terms of the power series for f(x)=3xcos(2x5)−6 about x=0
Expand Cosine Taylor Series: To find the power series for f(x)=3xcos(2x5)−6 about x=0, we need to expand the cosine function into its Taylor series and then multiply by 3x and subtract 6. The Taylor series for cos(z) about z=0 is 1−2!z2+4!z4−6!z6+…, where z is the input to the cosine function. In our case, z=2x5.
Simplify Series Terms: First, we write down the Taylor series expansion for cos(2x5) about x=0:cos(2x5)=1−2!(2x5)2+4!(2x5)4−6!(2x5)6+…
Multiply by 3x: Now, we simplify the terms of the series: cos(2x5)=1−24x10+2416x20−72064x30+…
Simplify Coefficients: Next, we multiply each term by 3x to get the series for 3xcos(2x5):3xcos(2x5)=3x−212x11+2448x21−720192x31+…
Subtract 6: We simplify the coefficients:3xcos(2x5)=3x−6x11+2x21−720192x31+…
Identify Non-Zero Term: Now, we subtract 6 from the series to get the final series for f(x):f(x)=3xcos(2x5)−6=(3x−6)−6x11+2x21−(720192)x31+…
Final Power Series Terms: We notice that the constant term −6 cancels with the constant term from 3x when x=0, so the first non-zero term is actually the term with x:f(x)=3x−6x11+2x21−(720192)x31+…
Final Power Series Terms: We notice that the constant term −6 cancels with the constant term from 3x when x=0, so the first non-zero term is actually the term with x:f(x)=3x−6x11+2x21−(720192)x31+...The first four non-zero terms of the power series for f(x) are:3x,−6x11,2x21, and −(720192)x31.
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