Add a Sequences and Series chapter to the Calculus course

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content/courses/calculus/differential/lines-of-tangency/index.md

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+---
+index: 9.2
+title: Tangent Lines
+type: lesson
+
+---
+
+In the previous lesson, we saw how derivatives can help us find the slope of a function. When the function's slope varies, the derivative gives us a formula for the slope, given the location.
+
+In this lesson, we'll look at an interesting application of the above — constructing lines tangent to nonlinear functions.
+
+<iframe src="https://www.desmos.com/calculator/vcpwd6c0uz?embed"  class="graph"></iframe>
+
+Let's say we have a function $f(x)$, and we want to construct a line tangent to the function at a point $\left(a,f(a)\right).$
+
+What should be the slope of the line? 
+
+For the function to 'align' with the line at the desired point, they must have the same slope. So, the slope of the line must be equal to $f'(a)$, the slope of the function at $x=a$.
+
+Using point-slope form, we have:
+$$\text{point} = (a, f(a))$$ $$\text{slope} = f'(a)$$
+
+$$y - f(a) = f'(a) (x-a).$$
+
+Now, we'll try this with an example.
+Let's say we want to graph a line tangent to $$y=x^3$$ at $$x=-1.$$
+
+For context, $y=x^3$ looks like this:
+
+<img class="graph" src="/img/graphs/x-cubed.png"/>
+
+Now, we'll use the point-slope approach we just came up with.
+
+To find the slope of $f(x)=x^3$, we need its derivative. Using the [power rule](/learn/calculus/differential/power-rule), we know that $$f'(x)=3x^2.$$
+So, the slope at $x=-1$ is $$f'(-1)=3(-1)^2=\boxed{3}.$$
+
+Using point-slope, we get  
+$$ y - (-1) = 3(x- (-1))$$ $$y+1=3x+3$$ $$\boxed{y=3x+2}.$$
+
+Now, try another example on your own. If you get stuck, use a hint!
+
+@1
+
+@2
+
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content/courses/calculus/differential/lines-of-tangency/problems/1.md

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+---
+type: problem
+answers:
+	- "x-\frac{\\pi}{2}"
+
+---
+
+@statement
+Find the equation of the line tangent to $y = \cos x$ at $x=\frac{\pi}{2}$. Express your answer in slope-intercept form.
+
+@hint
+What is the point of tangency?
+
+@hint
+What is the  slope at the point of tangency?
+
+@solution
+First, we identify the point of tangency. The question tells us to find the tangent line at $x=\frac{\pi}{2}.$ 
+So, our point is $$\left(\frac{\pi}{2}, \cos\left(\frac{\pi}{2}\right)\right)\rightarrow\left(\frac{\pi}{2}, 0\right).$$
+
+Now, we need the slope. The slope of $\cos x$ is modeled by its derivative, $\sin x$. 
+
+So, $$\cos'\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) = -1.$$
+
+Using point-slope, we get $$y-0=-1\cdot\left(x-\frac{\pi}{2}\right)$$   $$\boxed{y=\frac{\pi}{2}-x}.$$
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content/courses/calculus/differential/lines-of-tangency/problems/2.md

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+---
+type: problem
+answers:
+	- "8"
+
+---
+
+@statement
+If $f\left(x\right)=4-x^2,$ then the tangent lines at $x=-2$ and $x=2$ intersect at $\left(0,c\right).$ Find $c.$
+
+@hint
+How can we  break this problem into smaller parts? How many tangent lines do we need to find? 
+
+@hint
+Divide  the problem into smaller ones:
+1. Find the tangent line at $x=-2$
+2. Find the tangent line at $x=2$
+3. Find their intersection point
+
+@solution
+
+*For this problem, we will be needing the derivative of $f(x)$, so it is helpful to compute it ahead of time:* $$f'(x) = -2x.$$
+
+1. Find the tangent line at $x=-2:$
+
+$$\text{Point: } (-2,0)$$ $$\text{Slope: } f'(-2)=4$$ $$\text{Point-slope: } y-0=4(x+2)$$ $$y=4x+8.$$
+
+2. Find  the tangent line at $x=2:$
+	$$\text{Point: } (2,0)$$ $$\text{Slope: } f'(2)=4$$ $$\text{Point-slope: } y-0=-4(x-2)$$ $$y=8-4x.$$
+
+3. Find the intersection point:
+We know  that the $x$ value at the intersection point is $0$, so we just need to plug $x=0$ into one of  the equations.
+$$y=4x+8=0+8=\boxed{8}.$$
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content/courses/calculus/sequences-and-series/index.md

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+---
+index: 13.1
+title: Sequences and Series
+type: chapter
+
+---
+
+Coming soon!
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content/courses/calculus/sequences-and-series/mc-laurin/index.md

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+---
+index: 3.4
+title: MacLaurin Series
+type: lesson
+
+---
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content/courses/calculus/sequences-and-series/taylor-practice/index.md

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+---
+index: 3.3
+title: Practice with Taylor Series
+type: lesson
+
+---
+
+<!--stackedit_data:
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+-->

content/courses/calculus/sequences-and-series/taylor-practice/problems/1.md

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+---
+type: problem
+
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content/courses/calculus/sequences-and-series/taylor-series/index.md

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+---
+index: 3.2
+title: Introduction to Taylor Series
+type: lesson
+
+---
+
+In [Differential Calculus](/learn/calculus/differential), we learned that the derivative of a function describes its behavior at a specific point.
+
+We can use this information to construct a line tangent to the function. These tangent lines are often a good approximation for values close to the tangent point as well.
+
+But often, a line isn't a great representation of a function. Especially when the function isn't a line. The problem is, lines are always straight. They can tell us what direction a function is traveling in that moment, but it can't tell us where the function  had lunch today, its general mood, and where it plans to have dinner tonight.
+
+We can't achieve this with a straight line. But, what if we tried a curved line? Taking it up a degree from the line, let's see what we can do with a quadratic.
+
+As an example, we'll try to find a "tangent" quadratic to the  function $f(x) = \sin(x).$
+
+## Using Derivatives to Mimic Behavior
+
+Let's say we have a function $f(x)$, and we want to construct a tangent line at $x=a$.
+
+### Why do lines work?
+When we construct a line tangent to a function, we measure the slope (derivative) of the function at the tangent point, and assign that value to the slope of the line. In this way, the line and the function have the same first derivative at $x=a$.
+
+### Taking it a slope further
+The second derivative of every line is 0, unlike most functions you'll want to mimic. But, with a quadratic equation, we can control the second derivative.
+
+Let's find a quadratic $g(x)$ tangent to $f(x)$ at $x=a$.
+As opposed to the line, where we had two requirements, we now have three!
+They are: 
+$$f(a)=g(a)$$ $$f'(a)=g'(a)$$ $$f''(a)=g''(a).$$
+
+Take a moment to think about this yourself.
+
+---
+
+A naive  approach to this is
+$$g(x) = f(a) + f'(a)(x-a) + \frac{1}{2}f''(a)x^2.$$ 
+This only satisfies the third criteria.  If we take away the third term, it satisfies only the second criteria. If we take away the second term as well, it only satisfies the first criteria.
+
+In short, the higher degree erms are hindering the lower degree terms from serving their purpose.
+
+We can solve this by setting up higher degree terms to cancel out when they are not needed. This is done by replacing
+$$x \rightarrow x-a.$$
+
+Since these two terms have the same derivative, this replacement won't disturb the parts of our naive solution that work.
+
+So, we have
+$$g(x)=f(a)+f'(a)(x-a) + \frac{1}{2}f''(a)(x-a)^2.$$
+
+Let's try it out! With $f(x) = \cos x$ and $a=0$,  we have
+<img class="graph" src="/img/graphs/cosx-taylor-2.png" />
+
+This is definitely nicer than a line. And, it lines up well at $x=0$, which was our $a$ value. Now that we've seen greatness, let's get greedy!
+
+What should the next term  be to make this a cubic function?
+At first glance, you might say
+$$\frac{1}{3}f'''(a)(x-a)^3.$$
+The purpose of that fractional coefficient is to make up for the scalar residue caused by differentiation. 
+However, the problem here is that when this term comes into play, it will have been differentiated *thrice*,  and we need to cancel out not just a scalar of $3$, but $3!$. We didn't have to deal with this in the second  degree term, because $2!=2$. This factorial rule goes for any following terms.
+
+So, a cubic would look like
+$$g(x)=f(a)+f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2 + \frac{1}{3!}f'''(a)(x-a)^3.$$
+
+Sadly, in our example, the third degree term is $0$ because $\sin(0)=0.$ So, there's nothing new to see.  This doesn't satisfy our greed. Let's upgrade to the 8th degree!
+
+<img class="graph" src="/img/graphs/cosx-taylor-8.png" />
+
+That's more like it. And, if we graphed an infinite degree series, it would line up exactly!
+
+
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