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:

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

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 \(f'(x)\) 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, \(f'(x) = {e^{x + {x^2}}} \cdot (1 + 2x)\)
• Substitute \(x = 0\): \(f'(0) = {e^{0 + {0^2}}} \cdot (1 + 2 \cdot 0) = {e^0} \cdot (1) = 1 \cdot 1 = 1\)

Second Derivative Calculation

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

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

Third Derivative Calculation

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

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

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:

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

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

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.