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 Utilizando series conocidas determine la representacion en serie de potencias de ´
f(x) si:

The problem requires finding the power series representation of the function \( f(x) \) using known series expansions. The functions are given as:

\[
v) \quad f(x) = \int_0^1 e^{-x^2} - 1 \, dx
\]
\[
w) \quad f(x) = \int_0^x e^{-t^2} \, dt
\]

Let's begin with \( v) \) and \( w) \).

### For \( v) \):
We need to express \( f(x) = \int_0^1 e^{-x^2} - 1 \, dx \) using a series representation for \( e^{-x^2} \). The Taylor series for \( e^{-x^2} \) around \( x = 0 \) is:

\[
e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}
\]

Thus, we can rewrite the integrand \( e^{-x^2} - 1 \) as:

\[
e^{-x^2} - 1 = \sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}
\]

Now, we integrate term by term from 0 to 1:

\[
f(x) = \int_0^1 \sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!} \, dx
\]

This can be integrated for each term in the series.

### For \( w) \):
The function \( f(x) = \int_0^x e^{-t^2} \, dt \) involves the Gaussian integral, which does not have a simple closed-form solution. However, we can still expand \( e^{-t^2} \) in a power series as before:

\[
e^{-t^2} = \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{n!}
\]

Substituting this into the integral:

\[
f(x) = \int_0^x \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{n!} \, dt
\]

Now integrate term by term:

\[
f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_0^x t^{2n} \, dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \cdot \frac{x^{2n+1}}{2n+1}
\]

This is the power series representation for \( f(x) \).

---

Would you like a more detailed step-by-step solution for the integrals or clarifications on specific parts?

Here are some follow-up questions to consider:
1. Can you express the final power series for \( f(x) \) in a compact form?
2. How can you verify the convergence of the series representations?
3. What is the behavior of \( f(x) \) as \( x \to 0 \) and \( x \to \infty \)?
4. How does the series for \( e^{-x^2} \) compare to other known expansions, such as sine or cosine?
5. Can you use the power series to approximate \( f(x) \) for small \( x \)?

**Tip**: Power series can be a very effective way to approximate functions, especially when the exact integral is difficult to compute.

Utilizando series conocidas determine la representación en serie de potencias de f(x) si: v) f(x) = \int_0^1 e^{-x^2} - 1 \, dx y w) f(x) = \int_0^x e^{-t^2} \, dt.

Find the power series representation of f(x) using known series expansions for integrals involving e^(-x^2). This problem requires applying the Taylor series and integration techniques, suitable for university-level calculus students.

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