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Sam L.
MS in Math at UC Irvine, BA in Math at NYU, Former Math PhD Student, , Former Actuary
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Linear Algebra
TutorMe
Question:

Suppose U, V and W are all vector spaces where $$S: V \rightarrow W$$ and $$T: U \rightarrow V$$. Suppose also that $$S$$ and $$T$$ are linear maps. Show $$S \circ T$$ is a linear map.

Sam L.
Answer:

To show $$S \circ T$$ is a linear map we have to show it satisfies two properties. Note that a map $$f: A \rightarrow B$$ is linear if it satisfies: 1. $$f(x+y) = f(x)+f(y)$$, where $$x, y \in A$$. 2. $$f(c \cdot x) = c \cdot f(x)$$, where $$c$$ is a scalar. 1. Want to show: $$S \circ T(x+y) = S \circ T(x) + S \circ T(y)$$, where $$x, y \in U$$ To make this easier to work with let's put this composition notation in a more familiar form. So we want to show $$S(T(x+y)) = S(T(x)) + S(T(y))$$. We know by assumption that $$T$$ is a linear map, so $$T(x+y) = T(x)+T(y)$$. We now have $$S(T(x+y)) = S(T(x)+T(y))$$. Notice that by assumption, $$S$$ is also a linear map. Keep in mind that $$T(x)$$ and $$T(y)$$ are elements of $$V$$ so our domain for $$S$$ is still correct. So we have by linearity of $$S$$ that $$S(T(x)+T(y)) = S(T(x))+S(T(y)) = S \circ T(x) + S \circ T(y)$$, which is exactly what we wanted. 2. Want to show: $$S \circ T(c \dot x) = c \cdot S \circ T(x)$$, where $$c$$ is a scalar and $$x \in U$$. Again, let's make this a little easier to understand. $$S \circ T(c \cdot x) = S(T(c \cdot x))$$. By assumption we know $$T$$ is a linear map so $$T(c \cdot x) = c \cdot T(x)$$. This means $$S(T(c \cdot x))) = S(c \cdot T(x))$$. Again, $$T(x)$$ is in $$V$$ which is the correct domain for $$S$$. So we can continue. By assumption, $$S$$ is also a linear map, which means $$S(c \cdot T(x)) = c \cdot S(T(x)) = c \cdot S \circ T(x)$$, which is exactly what we wanted. So by proving this composition map satisfies the above two properties, we have show that $$S \circ T$$ is indeed a linear map.

Calculus
TutorMe
Question:

Find the derivative of $$\sin(\sqrt{x})$$.

Sam L.
Answer:

Let's assume we already know the derivative of $$\sin(x)$$ is $$\cos(x)$$. Unfortunately we can't just stop here on this problem because on the inside of the sine function, we don't just have $$x$$, but rather we have $$\sqrt{x}$$ which we have to account for. When you have a function inside another function, there's a good chance we will have to consider using the Chain Rule. In general it can be written out like $$(f(g(x)) )' = f'(g(x)) \cdot g'(x)$$. Here the function g(x) is inside the function f(x). In our problem, we have a function $$g(x) = \sqrt{x}$$ which is inside the function $$f(x)=\sin(x)$$. So $$f(g(x)) = \sin(\sqrt{x})$$. To compute its derivative, we need to find $$f'(g(x))$$ and $$g'(x)$$. The former is just finding the derivative of sine and plugging in g(x) i.e. $$f'(g(x)) = \cos(\sqrt{x})$$. $$g'(x)$$ will be $$(\sqrt{x})' = (x^{\frac{1}{2}})' = \frac{1}{2\sqrt{x}}$$ by the power rule. So according to our Chain Rule, $$(\sin(\sqrt{x}))' = \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$$.

Algebra
TutorMe
Question:

Find the solutions of the equation $$x^2 +kx +4=0$$.

Sam L.
Answer:

Usually the easiest way to find the solutions to this equation is by factoring. But, since we don't know what k is, this makes the problem a little harder. Remembering our quadratic formula we get $$x=\frac{-k \pm \sqrt{k^2-4(1)(4)}}{2}$$. So for whatever value k is, these two x-values $$x=\frac{-k + \sqrt{k^2-4(1)(4)}}{2}$$ and $$x=\frac{-k - \sqrt{k^2-4(1)(4)}}{2}$$ are the solutions to the above equation.

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