--- # System prepended metadata title: Week 8 --- # 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. ![image](https://hackmd.io/_uploads/ByDdLUe1bg.png) * A zero derivative can also be an infelction point which means it is neither maximum nor minimum. For eg $x^3$ at $x =0$. ![image](https://hackmd.io/_uploads/ryyIwUxJ-x.png) ### 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. ![image](https://hackmd.io/_uploads/By3gJOek-e.png). * 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): ![image](https://hackmd.io/_uploads/rkFeYOe1We.png) ### 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): ![image](https://hackmd.io/_uploads/rkUkINWybg.png) * 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: $$sin(x)= x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$ ![image](https://hackmd.io/_uploads/rJZk-Bb1Wg.png) * 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: ![image](https://hackmd.io/_uploads/HklGUHbJZg.png)