The image contains Taylor series expressions for Exercises 3 and 4, which involve constructing Taylor polynomials $T_n(f(x))$. Let's break them down and solve each one.

Exercise 3:

The given expression is:

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

Explanation:

• This is the Taylor series expansion for $a^x$ around $x = 0$ (Maclaurin series).
• $a^x$ can be written as $e^{x \log a}$, and the series expansion of $e^x$ is:

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

• Substituting $x \log a$ for $x$ in the expansion of $e^x$:

$e^{x \log a} = \sum_{k=0}^{n} \frac{(\log a)^k}{k!} x^k$

Thus, the series for $T_n(a^x)$ is:

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

This is the answer for Exercise 3.

Exercise 4:

The given expression is:

$T_n \left( \frac{1}{1+x} \right) = \sum_{k=0}^{n} (-1)^k x^k$

Explanation:

• The function $\frac{1}{1+x}$ is the classic geometric series expansion.
• The Taylor series for $\frac{1}{1+x}$ around $x = 0$ is:

$\frac{1}{1+x} = \sum_{k=0}^{\infty} (-1)^k x^k$

This is valid for $|x| < 1$.

• Since we are only considering the first $n$ terms, the Taylor polynomial is:

$T_n \left( \frac{1}{1+x} \right) = \sum_{k=0}^{n} (-1)^k x^k$

This is the answer for Exercise 4.

Final Answers:

1. For Exercise 3: $T_n(a^x) = \sum_{k=0}^{n} \frac{(\log a)^k}{k!} x^k$

2. For Exercise 4: $T_n \left( \frac{1}{1+x} \right) = \sum_{k=0}^{n} (-1)^k x^k$

Let me know if you need more details or explanations for these problems!

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Here are 5 related questions for practice:

1. How would you find the Taylor series for $\sin(x)$?
2. What is the general form of the Taylor series for $e^x$?
3. How would the Taylor series change if you expanded $a^x$ around a point other than 0?
4. What is the radius of convergence for the series $\sum_{k=0}^{\infty} (-1)^k x^k$?
5. How would you find the remainder term in the Taylor series approximation for $\frac{1}{1+x}$?

Tip: When finding Taylor series, always check the radius of convergence to ensure that the series is valid in the given domain.