# Week 8

## Quiz Summary

### Question 1
**If the derivative of a function is zero at some point, then that point must be a local minimum.**

**Answer :** False
**Explaination :**
* The derivative of a function is $0$ only because the slope of the tangent at that point is $0$.
* There is no confirmation if that is point is a local maximum or local minimum.
* For example, the graph of $sin(x)$ is oscillating and has points indicating local maximum and local minimum.

* A zero derivative can also be an infelction point which means it is neither maximum nor minimum. For eg $x^3$ at $x =0$.

### Question 2
**The function $f$ is concave up at point $t=10$**

**Answer :** True
**Explaination :**
* The function is concave upwards, because we can see from the graph that the function's derivative is negative before $t=10$ and positive i.e increasing after $t=10$.
* We can also verify by caluclating the second derivative of function at $t=10$. If it is positive, function is concave upwards.
* First derivative:
$$f'(t) = \frac{7t^6}{100000} - \frac{5t^4}{500} + \frac{3t^2}{10}$$
$$= 0.00007t^6 - 0.01t^4 + 0.3t^2$$

* Second derivative:
$$f''(t) = \frac{42t^5}{100000} - \frac{20t^3}{500} + \frac{6t}{10}$$
$$= 0.00042t^5 - 0.04t^3 + 0.6t$$

* put $t=10$
$$f''(10) = 0.00042(10^5) - 0.04(10^3) + 0.6(10)$$
$$f''(10) = 8 > 0$$

* Hence, statement is true.

### Question 3
**The tolerance level in Newton's Method represents the minimum (and final) difference between the estimated root and the actual root.**

**Answer :** False
**Explaination :**
* The tolerance in newton's method measures how the estimates are changing between iterations.
* We stop when the following condition is met:
$$|x_n -x_n-1| < tol$$
* This is an indication thatthe numerical method has converged.
* Hence it does not show the difference between esimated and actual root.
* For example, if:
$$|x_n -x_n-1| < 10^{-4}$$
* We only that know our estimate is stable within $0.0001$ but the actual difference $|x_n - r|$ might still be $2 \ x \ 10^{-4}$ or smaller or larger.

### Question 4
**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
**Explaination :**
* We can see from the graph that it touches $y=0$ at $t=10$ with a local minimum and upward concavity.
* Which means the derivative of the function at this point is $0$ as shown by the red tangent line.
.
* Hence, we can see that root newton's method could also be found by derivating $f(t)$ at that point.

### Question 5
**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
**Explaination :**
* The taylor sseries is based on the local information i.e derivatives at a point.
* The further you move away from expansion point $a$, taylor series becomes less accurate.
* As seen from the graph, it matches the orignal function upto a certain point and after which it deviates sharply.
* Hence, taylor series is not accurate for all $x$.
* We can get better approximation by using higher degree terms and accuracy increases at $a$.For example, if we use $30$ terms, we get following graphs(blue being orignal function):

### Question 6
**Let c and r be some constant values (like, $3, 1.4, \frac{2}{3}$, etc). Then, the Taylor Series CANNOT be used to find the roots of the function $f(x)=ce^{rx}$ because it is not infinitely differentiable (i.e., every derivative has its own non-zero derivative).**

**Answer :** False
**Explaination :**
* The function $f(x)=ce^{rx}$ is infinitely differentialble for all $x$.
* Taylor series exists and converges.
$$f(x)=3e^{1.4x}=3 \sum_{n=0}^{\infty} \frac{(1.4x)^n}{n!}$$
* If we use the first 5 terms, we get(blue being orignal function):

* The roots can be found without any approximation.Since, the function has no real roots as the function approaches $0$ asymptotically, we don't need taylor series to find them.

### Question 7
**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
**Explaination :**
* Taylor series uses all derivatives, not just first.
* When we neglect higher orders, we are just truncating, not defining it.
* The sinocidal functions are indefinitely differentiable. For eg:
$f(x)=sin(x)$
$f'(x)=cos(x)$
$f''(x)=-sin(x)$ ...etc
* Taylor series expansion is:
$$sin(x)= x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$

* Every term comes from different higher order derivative.
* Inaccuracies are only caused by ignoring higher terms when we want to see the approximation but by definition, the series is infinite and is exact when converges.

### Question 8
**The Taylor Series approximation of a sinusoidal function, such as $f(x)=50sin⁡(x)cos⁡(2x)+50$, provides a local polynomial approximation that effectively captures the periodic behavior of the function in the vicinity of the central point, a **

**Answer :** True
**Explaination :**
* The taylor series approximation captures the local oscillatory wave like behavior of the function at point $a$.
* At this expansion point, taylor series polynomial closely matches the curvature of the function.
* It recreates a small segment of oscillation around $a$, and as we move away from that point, the approximation loses accuracy. We can confirm from the graph: