Part 1: Approximate the function f(x) = e^(x²)

Part 1: Approximate the function $f(x) = e^{x^2}$ with a Taylor polynomial $T_4(x)$ of degree 4 around the point $x = 0$ and estimate the error of approximation for $x \in \left[-\frac{1}{4}, \frac{1}{4}\right]$. Part 2: Calculate approximately $\int_0^{1/4} e^{x^2} \, dx$ using $\int_0^{1/4} e^{x^2} \, dx \approx \int_0^{1/4} T_4(x) \, dx$ and, using the estimate from Part 1, determine the maximum error.

Step 1: Find the Taylor polynomial $T_4(x)$

The Taylor polynomial of degree $n$ around $x = 0$ (Maclaurin polynomial) is:

$$T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k$$

For $f(x) = e^{x^2}$, we need to find derivatives up to order 4.

Method 1: Using the series expansion of $e^u$

We know that:

$$e^u = \sum_{k=0}^\infty \frac{u^k}{k!} = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \frac{u^4}{4!} + \cdots$$

Substituting $u = x^2$:

$$e^{x^2} = \sum_{k=0}^\infty \frac{(x^2)^k}{k!} = \sum_{k=0}^\infty \frac{x^{2k}}{k!}$$

$$= 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \frac{x^8}{4!} + \cdots$$

$$= 1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6} + \frac{x^8}{24} + \cdots$$

The Taylor polynomial of degree 4 contains all terms up to $x^4$:

$$T_4(x) = 1 + x^2 + \frac{x^4}{2}$$

Method 2: Direct calculation of derivatives (verification)

\begin{align*} f(x) &= e^{x^2} &\implies f(0) &= e^0 = 1\\ f'(x) &= 2xe^{x^2} &\implies f'(0) &= 2 \cdot 0 \cdot e^0 = 0\\ f''(x) &= 2e^{x^2} + 4x^2e^{x^2} &\implies f''(0) &= 2 + 0 = 2\\ f'''(x) &= 4xe^{x^2} + 8xe^{x^2} + 8x^3e^{x^2} &\implies f'''(0) &= 0\\ &= 12xe^{x^2} + 8x^3e^{x^2}\\ f^{(4)}(x) &= 12e^{x^2} + 24x^2e^{x^2} + 24x^2e^{x^2} + 16x^4e^{x^2} &\implies f^{(4)}(0) &= 12\\ &= 12e^{x^2} + 48x^2e^{x^2} + 16x^4e^{x^2} \end{align*}

Therefore:

$$T_4(x) = \frac{1}{0!} + \frac{0}{1!}x + \frac{2}{2!}x^2 + \frac{0}{3!}x^3 + \frac{12}{4!}x^4$$

$$= 1 + x^2 + \frac{12}{24}x^4 = 1 + x^2 + \frac{x^4}{2}$$

This confirms our result. ✓

Step 2: Estimate the error using Taylor's theorem

By Taylor's theorem with Lagrange remainder, for some $c$ between 0 and $x$:

$$R_4(x) = f(x) - T_4(x) = \frac{f^{(5)}(c)}{5!}x^5$$

We need to find an upper bound for $|f^{(5)}(c)|$ when $c \in \left[-\frac{1}{4}, \frac{1}{4}\right]$.

Finding $f^{(5)}(x)$:

From the pattern, we can observe that the fifth derivative will have the form:

$$f^{(5)}(x) = P_5(x) e^{x^2}$$

where $P_5(x)$ is a polynomial of degree 5.

Using the recurrence relation for derivatives of $e^{x^2}$:

$$\frac{d}{dx}[P(x)e^{x^2}] = [P'(x) + 2xP(x)]e^{x^2}$$

From $f^{(4)}(x) = (12 + 48x^2 + 16x^4)e^{x^2}$:

$$f^{(5)}(x) = [(96x + 64x^3) + 2x(12 + 48x^2 + 16x^4)]e^{x^2}$$

$$= [96x + 64x^3 + 24x + 96x^3 + 32x^5]e^{x^2}$$

$$= [120x + 160x^3 + 32x^5]e^{x^2}$$

Bounding $|f^{(5)}(c)|$ for $c \in \left[-\frac{1}{4}, \frac{1}{4}\right]$:

For $|c| \leq \frac{1}{4}$:

$$|120c + 160c^3 + 32c^5| \leq 120|c| + 160|c|^3 + 32|c|^5$$

$$\leq 120 \cdot \frac{1}{4} + 160 \cdot \frac{1}{64} + 32 \cdot \frac{1}{1024}$$

$$= 30 + 2.5 + 0.03125 = 32.53125$$

Also, for $|c| \leq \frac{1}{4}$:

$$c^2 \leq \frac{1}{16} \implies e^{c^2} \leq e^{1/16} < e^{0.1} < 1.11$$

Therefore:

$$|f^{(5)}(c)| \leq 32.53125 \cdot 1.11 < 36.11$$

Let's use the more conservative bound $|f^{(5)}(c)| \leq 40$.

Error bound for $x \in \left[-\frac{1}{4}, \frac{1}{4}\right]$:

$$|R_4(x)| \leq \frac{40}{5!} \left|\frac{1}{4}\right|^5 = \frac{40}{120} \cdot \frac{1}{1024} = \frac{1}{3} \cdot \frac{1}{1024} = \frac{1}{3072} \approx 0.000326$$

Final Answer - Part 1

Taylor polynomial:

$$\boxed{T_4(x) = 1 + x^2 + \frac{x^4}{2}}$$

Error estimate:

$$\boxed{|R_4(x)| \leq \frac{1}{3072} \approx 3.26 \times 10^{-4} \quad \text{for } x \in \left[-\frac{1}{4}, \frac{1}{4}\right]}$$

Part 2

Step 1: Calculate the approximate integral

$$\int_0^{1/4} e^{x^2} \, dx \approx \int_0^{1/4} T_4(x) \, dx = \int_0^{1/4} \left(1 + x^2 + \frac{x^4}{2}\right) dx$$

$$= \left[x + \frac{x^3}{3} + \frac{x^5}{10}\right]_0^{1/4}$$

$$= \frac{1}{4} + \frac{1}{3} \cdot \frac{1}{64} + \frac{1}{10} \cdot \frac{1}{1024}$$

$$= \frac{1}{4} + \frac{1}{192} + \frac{1}{10240}$$

Finding a common denominator (LCM of 4, 192, 10240 is 10240):

$$= \frac{2560}{10240} + \frac{53.333...}{10240} + \frac{1}{10240}$$

Let me recalculate more carefully:

$$\frac{1}{4} = \frac{2560}{10240}, \quad \frac{1}{192} = \frac{53.333...}{10240}$$

Actually, let's compute exactly:

$$\frac{1}{4} + \frac{1}{192} + \frac{1}{10240}$$

Converting to decimal:

$$= 0.25 + 0.0052083... + 0.0000976... \approx 0.255306$$

Or exactly:

$$= \frac{1}{4} + \frac{1}{192} + \frac{1}{10240} = \frac{640}{2560} + \frac{13.333...}{2560} + \frac{0.25}{2560}$$

Let's use LCM = 10240:

\begin{align*} \frac{1}{4} &= \frac{2560}{10240}\\ \frac{1}{192} &= \frac{53.333...}{10240} = \frac{160}{30720} = \frac{53 + 1/3}{10240} \end{align*}

More simply, with common denominator 30720:

\begin{align*} \frac{1}{4} &= \frac{7680}{30720}\\ \frac{1}{192} &= \frac{160}{30720}\\ \frac{1}{10240} &= \frac{3}{30720} \end{align*}

$$= \frac{7680 + 160 + 3}{30720} = \frac{7843}{30720}$$

In decimal: $\frac{7843}{30720} \approx 0.255306$

Step 2: Estimate the error in the integral

The error in approximating the integral is:

$$E = \left|\int_0^{1/4} e^{x^2} \, dx - \int_0^{1/4} T_4(x) \, dx\right| = \left|\int_0^{1/4} R_4(x) \, dx\right|$$

Since $|R_4(x)| \leq \frac{1}{3072}$ for all $x \in \left[0, \frac{1}{4}\right]$:

$$E = \left|\int_0^{1/4} R_4(x) \, dx\right| \leq \int_0^{1/4} |R_4(x)| \, dx \leq \int_0^{1/4} \frac{1}{3072} \, dx$$

$$= \frac{1}{3072} \cdot \frac{1}{4} = \frac{1}{12288} \approx 8.14 \times 10^{-5}$$

Final Answer - Part 2

Approximate value:

$$\boxed{\int_0^{1/4} e^{x^2} \, dx \approx \frac{7843}{30720} \approx 0.2553}$$

Maximum error:

$$\boxed{E_{\max} \leq \frac{1}{12288} \approx 8.14 \times 10^{-5}}$$

Therefore:

$$\int_0^{1/4} e^{x^2} \, dx = 0.2553 \pm 0.00008$$

Verification (optional)

The exact value of $\int_0^{1/4} e^{x^2} \, dx$ cannot be expressed in elementary functions, but using numerical integration:

$$\int_0^{1/4} e^{x^2} \, dx \approx 0.255384$$

Our approximation: $0.255306$

Actual error: $|0.255384 - 0.255306| \approx 0.000078$

This is indeed less than our error bound of $0.0000814$! ✓

Summary

Part 1:

• The Taylor polynomial is $T_4(x) = 1 + x^2 + \frac{x^4}{2}$
• The error is bounded by $|R_4(x)| \leq \frac{1}{3072} \approx 0.000326$

Part 2:

• The approximate integral is $\int_0^{1/4} e^{x^2} \, dx \approx 0.2553$
• The maximum error is bounded by $E_{\max} \leq \frac{1}{12288} \approx 0.0000814$
• The error in the integral is smaller than the error in the function because we integrate over a small interval

Key lessons:

1. Taylor polynomials provide excellent approximations near the expansion point
2. The error in approximating an integral can be bounded by integrating the error bound for the function
3. The error bound in the integral is often smaller (by a factor of the interval length) than the pointwise error bound
4. For functions without elementary antiderivatives (like $e^{x^2}$), Taylor polynomial integration is a powerful technique