Let $f\left(x\right)=e^{x+x^2}$ for real x. From among the following, choose the Taylor series approximation of f(x) around x = 0, which includes all powers of x less than or equal to 3.

Taylor Series Approximation Explained

To find the Taylor series approximation of a function $f(x)$ around $x=0$ (also known as the Maclaurin series), we use the formula:

$f(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3+\dots$

The question asks for the approximation of $f(x)=e^{x+x^2}$ up to the powers of $x$ less than or equal to 3. This means we need to calculate the function value and its first three derivatives at $x=0$.

Function Value at x=0

First, we find the value of the function $f(x)$ at $x=0$:

- Given $f(x)=e^{x+x^2}$
- Substitute $x=0$: $f(0)=e^{0+0^2}=e^0=1$

First Derivative Calculation

Next, we calculate the first derivative $f^{\prime }(x)$ and evaluate it at $x=0$.

- We use the chain rule: If $y=e^u$, then $\frac{dy}{dx}=e^u\frac{du}{dx}$.
- Here, $u=x+x^2$, so $\frac{du}{dx}=1+2x$.
- Therefore, $f^{\prime }(x)=e^{x+x^2}\cdot (1+2x)$
- Substitute $x=0$: $f^{\prime }(0)=e^{0+0^2}\cdot (1+2\cdot 0)=e^0\cdot (1)=1\cdot 1=1$

Second Derivative Calculation

Now, we find the second derivative $f^{\prime \prime }(x)$ and its value at $x=0$. We will use the product rule $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$ for $f^{\prime }(x)=e^{x+x^2}\cdot (1+2x)$.

- Let $u=e^{x+x^2}$ and $v=(1+2x)$.
- We already found $u^{\prime }=e^{x+x^2}(1+2x)$.
- The derivative of $v$ is $v^{\prime }=2$.
- So, $f^{\prime \prime }(x)=u^{\prime }v+u{v}^{\prime }=\left(e^{x+x^2}(1+2x)\right)(1+2x)+e^{x+x^2}(2)$
- Simplify: $f^{\prime \prime }(x)=e^{x+x^2}(1+2x{)}^2+2e^{x+x^2}$
- Factor out $e^{x+x^2}$: $f^{\prime \prime }(x)=e^{x+x^2}\left[(1+2x{)}^2+2\right]$
- Substitute $x=0$: $f^{\prime \prime }(0)=e^{0+0^2}\left[(1+2\cdot 0{)}^2+2\right]=e^0\left[1^2+2\right]=1\cdot [1+2]=3$

Third Derivative Calculation

Finally, we calculate the third derivative $f^{\prime \prime \prime }(x)$ and its value at $x=0$. We use the product rule again for $f^{\prime \prime }(x)=e^{x+x^2}\left[(1+2x{)}^2+2\right]$.

- Let $A=e^{x+x^2}$ and $B=(1+2x{)}^2+2$.
- We know $A^{\prime }=e^{x+x^2}(1+2x)$.
- To find $B^{\prime }$, we differentiate $(1+2x{)}^2+2$: $B^{\prime }=2(1+2x)\cdot 2+0=4(1+2x)$.
- So, $f^{\prime \prime \prime }(x)=A^{\prime }B+A{B}^{\prime }=\left(e^{x+x^2}(1+2x)\right)\left[(1+2x{)}^2+2\right]+e^{x+x^2}[4(1+2x)]$
- Factor out $e^{x+x^2}(1+2x)$: $f^{\prime \prime \prime }(x)=e^{x+x^2}(1+2x)\left[(1+2x{)}^2+2+4\right]$
- Simplify: $f^{\prime \prime \prime }(x)=e^{x+x^2}(1+2x)\left[(1+2x{)}^2+6\right]$
- Substitute $x=0$: $f^{\prime \prime \prime }(0)=e^{0+0^2}(1+2\cdot 0)\left[(1+2\cdot 0{)}^2+6\right]$
- $f^{\prime \prime \prime }(0)=e^0(1)\left[1^2+6\right]=1\cdot 1\cdot [1+6]=7$

Summary of Values at x=0

Let's summarize the calculated values:Constructing the Taylor Series Approximation

Now, substitute these values into the Taylor series formula:

$f(x)=f(0)+f^{\prime }(0)x+\frac{f^{\prime \prime }(0)}{2!}{x}^2+\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3+\dots$

- Term for $x^0$: $f(0)=1$
- Term for $x^1$: $f^{\prime }(0)x=1\cdot x=x$
- Term for $x^2$: $\frac{f^{\prime \prime }(0)}{2!}{x}^2=\frac{3}{2\cdot 1}{x}^2=\frac{3}{2}{x}^2$
- Term for $x^3$: $\frac{f^{\prime \prime \prime }(0)}{3!}{x}^3=\frac{7}{3\cdot 2\cdot 1}{x}^3=\frac{7}{6}{x}^3$

Combining these terms, the Taylor series approximation of $f(x)=e^{x+x^2}$ around $x=0$ up to $x^3$ is:

$f(x)=1+x+\frac{3}{2}{x}^2+\frac{7}{6}{x}^3$

Alternative Method: Using Known Series Expansion

We can also use the known Maclaurin series for $e^u=1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\dots$

Let $u=x+x^2$. Substitute this into the series expansion for $e^u$:

$e^{x+x^2}=1+(x+x^2)+\frac{(x+x^2{)}^2}{2!}+\frac{(x+x^2{)}^3}{3!}+\dots$

Now, we expand each term and collect coefficients for powers of $x$ up to $x^3$:

- Constant Term: $1$
- **Term for $x$:** From $(x+x^2)$: $x$
- **Term for $x^2$:** From $(x+x^2)$: $x^2$ From $\frac{(x+x^2{)}^2}{2!}$: $\frac{1}{2}(x^2+2x^3+x^4)$. The $x^2$ term is $\frac{x^2}{2}$. Total $x^2$ term: $x^2+\frac{x^2}{2}=\frac{2x^2+x^2}{2}=\frac{3}{2}{x}^2$
- **Term for $x^3$:** From $\frac{(x+x^2{)}^2}{2!}$: $\frac{1}{2}(x^2+2x^3+x^4)$. The $x^3$ term is $\frac{2x^3}{2}=x^3$. From $\frac{(x+x^2{)}^3}{3!}$: $\frac{1}{6}{x}^3(1+x{)}^3=\frac{1}{6}{x}^3(1+3x+3x^2+x^3)$. The $x^3$ term is $\frac{x^3}{6}$. Total $x^3$ term: $x^3+\frac{x^3}{6}=\frac{6x^3+x^3}{6}=\frac{7}{6}{x}^3$

Combining these collected terms, we get the Taylor series approximation:

$1+x+\frac{3}{2}{x}^2+\frac{7}{6}{x}^3$

Both methods confirm the result.