Questions: Use the Taylor series shown in the table to find the first four nonzero terms of the Taylor series for the following function centered at 0 sin(6x^4) The first nonzero term of the Taylor series is

Solution Steps

To find the first four nonzero terms of the Taylor series for \(\sin(6x^4)\) centered at 0, we can use the Taylor series expansion for \(\sin(x)\), which is \(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}\). We substitute \(6x^4\) for \(x\) in this series and then expand it to find the first four nonzero terms.

The Taylor series expansion for \(\sin(x)\) around \(0\) is given by: \[ \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \] Substituting \(x\) with \(6x^4\), we have: \[ \sin(6x^4) = \sum_{n=0}^{\infty} \frac{(-1)^n (6x^4)^{2n+1}}{(2n+1)!} \]

Calculating the first few terms of the series, we find:

• For \(n=0\): \(\frac{(6x^4)^{1}}{1!} = 6x^4\)
• For \(n=1\): \(\frac{-(6x^4)^{3}}{3!} = -\frac{216x^{12}}{6} = -36x^{12}\)
• Higher-order terms will contribute to \(O(x^{16})\) and beyond.

Thus, the first four nonzero terms of the Taylor series for \(\sin(6x^4)\) are: \[ 6x^4, -36x^{12}, O(x^{16}) \]

Final Answer