The Taylor series about x=a for a function is given by

\(\displaystyle \sum_{n=0}^{\infty}f^{(n)}(a)\frac{(x-a)^n}{n!}\)

So, for the first three terms (n=0, 1, 2), we must find the zeroth, first, and second derivatives. The zeroth derivative is just the function itself.

\(\displaystyle f'(x)=3e^{3x}-\sin(x)\)

\(\displaystyle f''(x)=9e^{3x}-\cos(x)\)

The derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^{ax}=ae^{ax}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)\)

Now, using the above formula, we can write out the first three terms:

\(\displaystyle (e^{6\pi}+1)\frac{(x-2\pi)^0}{0!}+3e^{6\pi}\frac{(x-2\pi)^1}{1!}+(9e^{6\pi}-1)\frac{(x-2\pi)^2}{2!}\)

which simplifies to

\(\displaystyle (e^{6\pi}+1)+3e^{6\pi}(x-2\pi)+\frac{(9e^{6\pi}-1)(x-2\pi)^2}{2}\)

The formula for the Maclaurin Series of a function is defined as follows

\(\displaystyle P(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)x^n}{n!}\) where \(\displaystyle f^{(n)}(0)\) is the n-th derivative when \(\displaystyle x=0\)

For the third degree polynomial we solve find the sum with the upper bound \(\displaystyle n=3\)

\(\displaystyle \sum_{n=0}^{3}\frac{f^{(n)}(0)x^n}{n!}\)

First we evaluate 

\(\displaystyle f(0)=e^{3(0)}=1\)

We must also solve for the first, second, and third derivative of \(\displaystyle f(x)\)

\(\displaystyle f'(x)=3e^{3x}\)

\(\displaystyle f''(x)=9e^{3x}\)

\(\displaystyle f^{(3)}(x)=27e^{3x}\)

When \(\displaystyle x=0\) we find the derivative values of

\(\displaystyle f'(0)=3e^{3(0)}=3\)

\(\displaystyle f''(0)=9e^{3(0)}=9\)

\(\displaystyle f^{(3)}(0)=27e^{3(0)}=27\)

Using these values we find the third degree Maclaurin Polynomial to be

\(\displaystyle P(x)=1+3x+\frac{9x^2}{2}+\frac{27x^3}{3!}\)

The Taylor series about x=a for any function is given by

\(\displaystyle \sum_{n=0}^{\infty}\frac{f^{n}(a)(x-a)^n}{n!}\)

We must take the nth derivative of the given function and determine the trend as n goes to infinity. To determine this, we start at n=0 (the zeroth derivative, or the function itself), and go further:

\(\displaystyle f^0(x)=x\ln x\)

\(\displaystyle f'(x)=\ln(x)+\frac{x}{x}=\ln(x)+1\)

\(\displaystyle f''(x)=\frac{1}{x}\)

\(\displaystyle f'''(x)=-\frac{1}{x^2}\)

\(\displaystyle f^{(4)}(x)=2x^{-3}\)

\(\displaystyle f^{(5)}(x)=-6x^{-4}\)

The derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)\cdot f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\ln(x)=\frac{1}{x}\)

We must now find a pattern for the derivatives we took. The zeroth and first derivatives do not follow a pattern, but the derivatives after do follow a pattern: the sign alternates, the power of x decreases by 1, and the coefficient is the factorial of \(\displaystyle (n-2)\) starting at the n=2 derivative. 

The derivative can then be expressed as 

\(\displaystyle f^{(n)}(a)=(-1)^n(n-2)!(x)^{-n+1}\)

for n greater than or equal to 2.

For the Taylor series itself, we must evaluate the derivative at x=a=7. When we do this, and write the pattern elements for the derivatives past n=1, we get

\(\displaystyle 7\ln 7+\ln7 +1 + \sum_{n=0}^{\infty} (-1)^n(n-2)(n-1)(7)^{-n+1}(x-7)^n\)

To obtain the Taylor Series for \(\displaystyle e^{2x}\), we start with the Taylor Series for \(\displaystyle e^{x} = \sum_{n=0}^\infty \frac{x^n}{n!}\), and substitute \(\displaystyle x\) on both sides with \(\displaystyle 2x\), and simplify.

\(\displaystyle e^{2x} = \sum_{n=0}^\infty \frac{(2x)^n}{n!} = \sum_{n =0}^\infty \frac{2^nx^n}{n!}=\sum_{n=0}^\infty2^n\frac{x^n}{n!}\).

To obtain the Taylor Series for \(\displaystyle x^2\sin(x^2)\), we start with the Taylor Series for \(\displaystyle \sin(x) = \sum_{n=0}^\infty \frac{(-1)^nx^{n+1}}{(n+1)!}\).

\(\displaystyle \sin(x^2) = \sum_{n=0}^\infty \frac{(-1)^n(x^2)^{n+1}}{(n+1)!}\). (Substitute \(\displaystyle x\) with \(\displaystyle x^2\) on both sides)

\(\displaystyle x^2\sin(x^2) = x^2\sum_{n=0}^\infty \frac{(-1)^n(x^2)^{n+1}}{(n+1)!}\). (Multiply both sides by \(\displaystyle x^2\),

\(\displaystyle = \sum_{n=0}^\infty x^2\frac{(-1)^nx^{2n+2}}{(n+1)!} =\sum_{n=0}^\infty\frac{(-1)^nx^{2n+4}}{(n+1)!}\).

We can write \(\displaystyle y\) as a Mclaurin series by saying that:

\(\displaystyle y=\sum_{0}^{\infty} \frac{y^{(n)}(0)}{n!}(x)^n\)

Since y is just a polynomial expression:

\(\displaystyle y''=\sum_{2}^{\infty} (n)(n-1)\frac{y^{(n)}(0)}{n!}(x)^{n-2}\)

The reason we must bring the lower bound to 2 is because Taylor Series can ONLY be written as a summation of polynomials and no rational components. 

The Maclaurin series for \(\displaystyle e^{x}\), taken to the third term, is:

\(\displaystyle e ^{x} = 1 + x + \frac{x^{2}}{2}+...\)

Substitute \(\displaystyle 2x^{2}\) for \(\displaystyle x\):

\(\displaystyle f(x) = e ^{2x ^{2}}\)

\(\displaystyle e ^{x} = 1 + 2x^{2} + \frac{(2x^{2})^{2}}{2}+...\)

\(\displaystyle e ^{x} = 1 + 2x^{2} + \frac{4x^{4}}{2}+...\)

\(\displaystyle e ^{x} = 1 + 2x^{2} + 2x^{4} +...\)

Rewrite this function as 

\(\displaystyle f(x) = \sqrt{3^x} =\left ( \sqrt{3} \right )^x\)

\(\displaystyle = \left ( 3 ^{\frac{1}{2}} \right )^x = \left ( e^{\ln 3 ^{\frac{1}{2}}} \right )^x = \left ( e^{\frac{1}{2} \ln 3} \right )^x= \left ( e^{\frac{1}{2} \ln 3 \; \cdot x } \right )\)

The Maclaurin series for \(\displaystyle e^{x}\), taken to the third term, is:

\(\displaystyle e ^{x} = 1 + x + \frac{x^{2}}{2}+...\)

Substitute \(\displaystyle \frac{1}{2} \ln 3 \; \cdot x\) for \(\displaystyle x\):

\(\displaystyle f(x) = \sqrt{3^x} = 1 + \frac{1}{2} \ln 3 \; \cdot x + \frac{(\frac{1}{2} \ln 3 \; \cdot x)^{2}}{2}+...\)\(\displaystyle = 1 + \frac{1}{2} \ln 3 \; \cdot x + \frac{\left \(\frac{1}{4} \right\)( \ln 3 )^{2} x^{2}}{2}+...\)

\(\displaystyle = 1 + \left ( \frac{\ln 3}{2} \right ) x +\left [ \frac{ ( \ln 3 )^{2} }{8} \right ] \; x^{2}+...\)

The Maclaurin series for \(\displaystyle \cos x\) is 

\(\displaystyle \cos x = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} -...\)

Substitute \(\displaystyle \sqrt{2x}\) for \(\displaystyle x\). The series becomes

\(\displaystyle \cos \sqrt{2x} = 1 - \frac{\left ( \sqrt{2x} \right )^{2}}{2} + \frac{\left ( \sqrt{2x} \right)^{4}}{24} -...\)

\(\displaystyle = 1 - \frac{2x}{2} + \frac{4x^{2}}{24} -...\)

\(\displaystyle = 1 - x + \frac{1}{6}x^{2} -...\)

The Maclaurin series for \(\displaystyle \cos x\) is 

\(\displaystyle \cos x = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} -...\)

Substitute \(\displaystyle 2x^{2}\) for \(\displaystyle x\). The series becomes

\(\displaystyle \cos \left (2x^{2} \right ) = 1 - \frac{\left (2x^{2} \right )^{2}}{2} + \frac{\left (2x^{2} \right )^{4}}{24} -...\)

\(\displaystyle = 1 - \frac{ 4x^{4} }{2} + \frac{ 16x^{8} }{24} -...\)

\(\displaystyle = 1 -2x^{4} + \frac{ 2 }{3}x^{8} -...\)