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# System prepended metadata

title: Week_8_writeup

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**Q1) If the derivative of a function is zero at some point, then that point must be a local minimum.**

**Answer:** False

**Explanation:** 
When the derivative of a function \( f(x) \) is zero at a point \( x = c \), it only means that \( x = c \) is a *critical point*. This does not necessarily mean \( x = c \) is a local minimum.

For example, consider the function 
$$
f(x) = x^3.
$$
The derivative of \( f(x) \) is 
$$
f'(x) = \frac{d}{dx} (x^3) = 3x^2.
$$
To find the critical points, we set \( f'(x) = 0 \):
$$
3x^2 = 0.
$$
Solving for \( x \), we get:
$$
x = 0.
$$
So, \( x = 0 \) is a critical point of \( f(x) \).

At \( x = 0 \), \( f'(0) = 0 \), so \( x = 0 \) is a critical point. However, if we check the behavior of \( f(x) \) around \( x = 0 \), we see that:
- \( f(x) < 0 \) for \( x < 0 \)
- \( f(x) > 0 \) for \( x > 0 \)

This shows that \( x = 0 \) is not a local minimum but an *inflection point*, where the function changes direction.

In general:
1. A point where \( f'(x) = 0 \) can be a local minimum, a local maximum, or neither.
2. To determine if it's a local minimum, we need additional tests, such as the *second derivative test*.

Thus, the statement is **False** because \( f'(x) = 0 \) alone does not guarantee a local minimum.



**Q2)The function is concave up at the point.**

**Answer:** True

**Explanation:** For a function to be concave up at a point, the second derivative at that point needs to be positive.

**Function Definition:**
   The function is:
   $$
   f(t) = \frac{t^7}{100000} - \frac{t^5}{500} + \frac{t^3}{10}.
   $$

**Second Derivative:**
   The second derivative of \( f(t) \) is:
   $$
   f''(t) = \frac{42t^5}{100000} - \frac{20t^3}{500} + \frac{6t}{10}.
   $$

**Check at \( t = 10 \):**
   To see if the function is concave up at \( t = 10 \), we substitute \( t = 10 \) into \( f''(t) \):
   
   $$
   f''(10) = \frac{42 \cdot 10^5}{100000} - \frac{20 \cdot 10^3}{500} + \frac{6 \cdot 10}{10}.
   $$

**Calculation:**
   - First term: $$ \frac{42 \cdot 100000}{100000} = 42 $$
   - Second term: $$ \frac{20 \cdot 1000}{500} = 40 $$
   - Third term: $$ \frac{6 \cdot 10}{10} = 6 $$

   So,
   $$
   f''(10) = 42 - 40 + 6 = 8.
   $$

   Since \( f''(10) = 8 \), which is positive, the function is concave up at \( t = 10 \). This means the statement is **True**.

**Q3) The tolerance level in Newton's Method represents the maximum (and final) difference between the estimated root and the actual root.**

**Answer:** False

**Explanation:** 

In Newton's Method, the tolerance level is used as a stopping point for the calculations, not as an exact measure of how close we are to the actual root. 

The tolerance level, represented by \( \epsilon \), is a small value we set so that when the difference between two guesses is smaller than \( \epsilon \), we stop the calculations. This difference is given by:
   $$
   |x_{n+1} - x_n| < \epsilon,
   $$
   where \( x_{n+1} \) and \( x_n \) are two consecutive guesses.

**Tolerance Does Not Guarantee the Exact Error:**
Just because the difference between guesses is within the tolerance, it doesn’t mean the estimated root is exactly within that tolerance of the actual root. The true difference between our guess and the actual root could still be larger or smaller.

Therefore, the tolerance level only tells us when to stop guessing, not the exact difference between the estimated root and the true root. This makes the statement **False**.

**Q4) In this case, the root that Newton's Method was attempting to approximate could have been calculated by finding a local minimum of $f$.**

**Answer:** True

**Explanation:** 

In this case, the root that Newton's Method was attempting to approximate could have been calculated by finding a local minimum of $f$" is **True** due to the following reasons:

1. A root $t_0$ of the function $f(t)$ is defined as a point where $f(t_0) = 0$.

2. The function is continuous and differentiable in the given interval.

3. A local minimum indicates that the function value is lower than its neighbors, which can suggest the behavior of the function near a root.

4. If $f(t)$ approaches zero near a local minimum, this indicates that the local minimum could be linked to the root of the function. If the local minimum is zero, it is directly a root.

5. Since Newton's Method iteratively approximates roots based on function values and their derivatives, starting close to a local minimum can enhance convergence to a nearby root.

6. Therefore, if a local minimum exists and its value is close to zero, Newton's Method could efficiently approximate this root.


**Q5) The Taylor Series approximation provides an exact representation of the original function $f(x)$ for all values of $x$ ensuring perfect accuracy in estimating website traffic for the upcoming week.**

**Answer:** False

**Explanation:**

The statement "The Taylor Series approximation provides an exact representation of the original function $f(x)$ for all values of $x$ ensuring perfect accuracy in estimating website traffic for the upcoming week" is **False** due to the following reasons:

1. The Taylor Series approximation is a polynomial expansion of a function around a specific point $a$. While it can provide a good approximation near this point, it does not guarantee accuracy for values of $x$ far from $a$.

2. The convergence of a Taylor Series depends on the function and the chosen expansion point. For some functions, the Taylor Series may converge to the function value only within a certain interval (the radius of convergence).

3. In the case of periodic functions like $f(x) = 50\sin(x)\cos(2x) + 50$, the Taylor Series cannot capture the periodic behavior accurately for all values of $x$. It becomes increasingly inaccurate as we move further away from the expansion point, resulting in a poor approximation.

4. The polynomial nature of the Taylor Series limits its ability to represent oscillatory behavior, such as the sine and cosine functions, leading to discrepancies in estimating the website traffic for days beyond a few days from the expansion point.

5. Therefore, while the Taylor Series provides a local approximation of $f(x)$, it cannot ensure perfect accuracy in estimating website traffic for the upcoming week, particularly for values of $x$ that are not close to the expansion point.

**Q6) A function containing $e^x$ is similar to our function in that it would be infinitely differentiable. (I.e., every derivative has its own non-zero derivative.**

**Answer:** True

**Explanation:**

The statement "A function containing $e^x$ is similar to our function in that it would be infinitely differentiable. (I.e., every derivative has its own non-zero derivative.)" is **True** for the following reasons:

1. **Infinitely Differentiable Functions**: A function is considered infinitely differentiable if it has derivatives of all orders. The exponential function $e^x$ is known for being infinitely differentiable over the entire set of real numbers.

2. **Derivative Properties**: The derivative of $e^x$ is $e^x$, which means that all derivatives of $e^x$ are equal to the original function. Therefore, each derivative is non-zero for any real value of $x$. This property ensures that the function does not flatten out or become zero at any point.

3. **Comparison with Trigonometric Functions**: Similar to the function $f(x) = 50\sin(x)\cos(2x) + 50$, which is also infinitely differentiable, functions that involve trigonometric expressions can be differentiated repeatedly without losing differentiability. Each derivative will also yield a sinusoidal function, maintaining the infinite differentiability.

4. **Convergence and Series Representation**: Both the function involving $e^x$ and the trigonometric function can be approximated using their respective Taylor Series. The Taylor Series for both functions converge to the original function for all values of $x$, further highlighting their infinitely differentiable nature.

Therefore, just like the function $f(x)$, any function containing $e^x$ is infinitely differentiable, and each of its derivatives remains non-zero, making the statement true.

**Q7) The Taylor Series expansion of a sinusoidal function involves computing only the function's first-order derivatives at the expansion point, neglecting higher-order derivatives, which may lead to inaccuracies in the polynomial approximation**

**Answer:** False

**Explanation:**

The statement "The Taylor Series expansion of a sinusoidal function involves computing only the function's first-order derivatives at the expansion point, neglecting higher-order derivatives, which may lead to inaccuracies in the polynomial approximation" is **False** for the following reasons:

1. **Taylor Series Definition**: The Taylor Series expansion of a function at a point $a$ includes terms involving all derivatives of the function at that point. Specifically, for a function $f(x)$, the Taylor Series is given by:
   $$
   f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots
   $$
   This shows that the Taylor Series considers not only the first-order derivative but all higher-order derivatives as well.

2. **Sinusoidal Functions**: For sinusoidal functions such as $\sin(x)$ and $\cos(x)$, the derivatives are periodic:
   - $f(x) = \sin(x)$ has derivatives $\cos(x)$, $-\sin(x)$, $-\cos(x)$, and $\sin(x)$ again, repeating every four derivatives.
   - This periodicity implies that all derivatives contribute to the Taylor Series expansion.

3. **Accuracy of the Approximation**: Neglecting higher-order derivatives would lead to an incomplete Taylor Series, resulting in a polynomial that may not accurately capture the behavior of the original sinusoidal function, especially for values of $x$ further away from the expansion point $a$.

4. **Convergence**: The Taylor Series converges to the original function in its vicinity, and including only the first-order derivative would not yield an accurate approximation. Higher-order derivatives become increasingly important as we expand the series, particularly for functions that oscillate, like sinusoidal functions.

Therefore, the statement is false because a proper Taylor Series expansion includes all derivatives, and neglecting higher-order derivatives can indeed lead to inaccuracies in the polynomial approximation.

**Q8) The Taylor Series expansion of a sinusoidal function, such as $$f(x) = 50\sin(x)\cos(2x) + 50$$ , provides a local polynomial approximation that effectively captures the periodic behavior of the function in the vicinity of the central point, $a$**

**Answer:** True

**Explanation:**

The statement "The Taylor Series expansion of a sinusoidal function, such as 
$$f(x) = 50\sin(x)\cos(2x) + 50$$
, provides a local polynomial approximation that effectively captures the periodic behavior of the function in the vicinity of the central point, $a$" is **True** for the following reasons:

1. **Definition of Taylor Series**: The Taylor Series expansion of a function around a point $a$ is given by:
   $$
   f(x) =  {\displaystyle f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3} + \cdots = \lim_{N\to \infty}\sum_{n=0}^{N}{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}.}
   $$

   $$
   50 + 50x - \frac{325 x^3}{3} + \frac{605 x^5}{12} - \frac{5465 x^7}{504} + \frac{49205 x^9}{36288} - \frac{88573 x^{11}}{798336} + \cdots
   $$
   This series provides a polynomial approximation of the function $f(x)$ near the point $a$.

2. **Local Approximation**: The Taylor Series effectively approximates the function locally. When $x$ is close to $a$, the Taylor polynomial can closely mimic the behavior of the original function, capturing key features such as the function's value and slope at that point.

3. **Periodic Nature of Sinusoidal Functions**: For the specific function $f(x) = 50\sin(x)\cos(2x) + 50$, the underlying sinusoidal components have derivatives that exhibit periodic behavior. The derivatives of sinusoidal functions repeat in a predictable manner, allowing the Taylor Series to accurately reflect this periodicity in its polynomial form.

4. **Higher-Order Terms**: Including higher-order terms in the Taylor Series expansion ensures that the approximation captures the oscillatory nature of sinusoidal functions. Each successive derivative provides additional information about the curvature and oscillation of the function, enhancing the polynomial's fidelity to the original function.

5. **Convergence Around the Center Point**: The Taylor Series converges to the original function as $x$ approaches $a$, making it a powerful tool for analyzing the function's behavior in that neighborhood. For sinusoidal functions, this means that the periodic behavior is effectively represented within a certain range around the central point.

Therefore, the statement is true because the Taylor Series expansion provides a local polynomial approximation that effectively captures the periodic behavior of the function $f(x)$ in the vicinity of the central point $a$.
