# Tutor profile: David T.

## Questions

### Subject: Discrete Math

Prove that $$1^{3} + 2^{3} + \ldots + n^{3} = \left[\frac{(n)(n+1)}{2}\right]^{2}$$ for all integers $$ n \geq 1$$.

There is an infinite number of integers $$n \geq 1$$, such that it would be impossible to check that they all work in this formula. Instead, in order to prove that this formula works for all of them, we use the process of induction. A proof by induction has two parts. First, we prove that the hypothesis is true for a base case, or if necessary, a few base cases. In this proof, we will show that for the integer $$n = 1$$, the formula given holds true. The second part of the proof requires an assumption that the hypothesis is true for some positive integer, which we will call $$k$$. This assumption is referred to as the inductive hypothesis. We then show that it must also be true for the number $$k+1$$ given our inductive hypothesis. This is enough to show the conjecture is true for all integers $$n \geq 1$$, because $$k$$ could have been $$1$$, which would mean that $$n = 2$$ is true. Similarly, $$k$$ could have been $$2$$ which would have meant that $$n = 3$$ is true, and so on. A full proof is given below. Proof: Base Case: Let $$n= 1$$. Then $$1^{3} = 1$$. Similarly, $$\left[\frac{(1)(2)}{2}\right]^{2} = 1^{2} = 1$$. The formula is true for $$n = 1$$. Inductive Hypothesis: Assume $$k \in \mathbb{Z}$$ such that $$1^{3} + 2^{3} + \ldots + k^{3} = \left[\frac{(k)(k+1)}{2}\right]^{2}$$. [We must show that $$1^{3} + 2^{3} + \ldots + k + (k+1)^{3} = \left[\frac{(k+1)(k+2)}{2}\right]^{2}$$] $$1^{3} + 2^{3} + \ldots + k + (k+1)^{3} = \left[\frac{(k)(k+1)}{2}\right]^{2} + (k+1)^{3}$$ by the Inductive Hypothesis. We can simplify the right side of the equation to get RHS = $$\left[\frac{k^{2}+k}{2}\right]^{2} + k^{3}+3k^{2}+3k+1$$ = $$\frac{k^{4}+2k^{3}+k^{2}}{4} + \frac{4k^{3}+12k^{2}+12k+4}{4}$$ = $$\frac{k^{4}+6k^{3}+13k^{2}+12k+4}{4}$$ So we have that $$1^{3} + 2^{3} + \ldots + k + (k+1)^{3} = \frac{k^{4}+6k^{3}+13k^{2}+12k+4}{4}$$ and we want to show that $$1^{3} + 2^{3} + \ldots + k + (k+1)^{3} = \left[\frac{(k+1)(k+2)}{2}\right]^{2}$$. But $$\left[\frac{(k+1)(k+2)}{2}\right]^{2} = \frac{k^{4}+3k^{3}+2k^{2}+3k^{3}+9k^{2}+6k+2k^{2}+6k+4}{4} = \frac{k^{4}+6k^{3}+13k^{2}+12k+4}{4}$$. So from our inductive hypothesis we have shown that $$1^{3} + 2^{3} + \ldots + k + (k+1)^{3} = \frac{k^{4}+6k^{3}+13k^{2}+12k+4}{4} = \left[\frac{(k+1)(k+2)}{2}\right]^{2}$$, so $$1^{3} + 2^{3} + \ldots + n^{3} = \left[\frac{(n)(n+1)}{2}\right]^{2}$$ for all integers $$ n \geq 1$$. $$\square$$

### Subject: Calculus

Let $$f(x)$$, $$g(x)$$, and $$h(x)$$ be continuous, differentiable functions. Find the derivative of $$f(g(h(x)))$$ in terms of $$f$$, $$g$$, $$h$$, and their derivatives. For further thought, consider how the derivative would change if the third function was $$h(2x)$$.

This problem is a simple application of the chain rule. The trick to it is that we aren't actually given the functions to work with, so we must really be sure we understand conceptually what is going on when we do the chain rule. We have a function $$h(x)$$, which is embedded into the function $$g(x)$$ and that whole composition is then plugged in as the input of the function $$f(x)$$. This means that $$f(x)$$ is what we could consider being the "outside" function. The chain rule tells us that the derivative of a composition of functions is the derivative of the outside function in which all of the inner parts remain unchanged, multiplied by the derivative of the inner parts. In this case, the inner parts would be $$g(h(x))$$. So, $$[f(g(h(x)))]' = [f'(g(h(x)))]*[g(h(x))]'$$ Now we still have$$[g(h(x))]'$$ that needs to be simplified, and once again, the chain rule applies. This time, the outside function is $$g(x)$$ and the inside function is $$h(x)$$, so by the chain rule, $$[g(h(x))]' = [g'(h(x))]*[h'(x)]$$. Combining these two equations together, we have, $$[f(g(h(x)))]' = [f'(g(h(x)))]*[g'(h(x))]*[h'(x)]$$. This is the solution to the initial problem. For the further thought question, we must consider $$2x$$ as a function of its own. First, let us call $$j(x) = 2x$$. Then our problem simply becomes finding the derivative of $$f(g(h(j(x))))$$, which follows the same process as the previous problem, but with an additional chain. The derivative becomes: $$[f(g(h(j(x))))]' = [f'(g(h(j(x))))]*[g'(h(j(x)))]*[h'(j(x))]*[j'(x)]$$ But, since $$j(x) = 2(x)$$ and the derivative $$j'(x) = 2$$ can easily be calculated, this can be simplified to get the solution: $$[f(g(h(2x)))]' = [f'(g(h(2x)))]*[g'(h(2x))]*[h'(2x)]*[2]$$.

### Subject: Algebra

Let $$x = 2+3i$$ be a root of the degree $$3$$ polynomial $$x^{3}-3x+52$$. What are the other roots of this polynomial?

This question wants us to find the roots of the polynomial $$x^{3}-3x+52$$. Since this polynomial is a degree $$3$$, we know that there will be $$3$$ roots. We already are given that $$x = 2+3i$$ is one of the roots, so we only have $$2$$ more to find. By the law of conjugates, we know that any polynomial with a root with an imaginary term (in this case $$x = 2+3i$$ has a $$3i$$, which is imaginary) also has its conjugate as a root. That is, the same exact root, but the sign of the imaginary term has been changed. By this fact, we know that $$x = 2-3i$$ is also a root of our polynomial. A polynomial can always be rewritten as the product of its factors, and the law of factors tells us that these factors can be determined by the roots of the polynomial. From this information, we know that $$(x-(2+3i))$$ and $$(x-(2-3i))$$ are factors of $$x^{3}-3x+52$$ with a third, unknown linear factor. We can set up an equation now to help us find what that factor is. We write: $$x^{3}-3x+52$$ = $$(x-(2+3i))(x-(2-3i))(x-c)$$ for some unknown value c. That number c is the value of the final root of the polynomial we are looking for. Now we just need to manipulate the equation to find what c is. First we distribute the negative sign in each of the known factors: $$x^{3}-3x+52$$ = $$(x-2-3i)(x-2+3i)(x-c)$$ Then we can multiply the two factors together, leaving the $$(x-c)$$ alone. Remember that when multiplying two factors together in which each has three terms, each term in the first factor must be multiplied by each term in the second factor. We get: $$x^{3}-3x+52$$ = $$(x^{2}-2x+3ix-2x+4-6i-3ix+6i-9i^{2})(x-c)$$ As will always be the case when multiplying factors that are formed from conjugates, all of the terms with $$i$$ in them cancel out nicely. Remember that $$i^{2}=-1$$. $$x^{3}-3x+52$$ = $$(x^{2}-4x+13)(x-c)$$ Now, to solve for the $$(x-c)$$ term, think of the entire factor as a variable. Just as if you were solving the equation $$10=5q$$, you would divide both sides of the equation by the 5 to isolate your unknown value. Similarly, here we will divide both sides of the equation by $$(x^{2}-4x+13)$$ to isolate $$(x-c)$$, remembering to use long division techniques. The first step for completing the long division is to fill in all missing powers of x with placeholders inside the division bar. This means, that we should now have $$x^{3}+0x^{2}-3x+52$$. Now we look at the first term on the outside and on the inside of the division bar. On the outside, we have $$x^{2}$$. On the inside, we have $$x^{3}$$. We must ask ourselves, what do we multiply the outside by to get the number on the inside? The value we multiply $$x^{2}$$ by to get $$x^{3}$$ is just $$x$$, so we write that number on top of the division bar. Then, we multiply the $$x$$ by $$x^{2}-4x+13$$ and write it below the $$x^{3}+0x^{2}-3x+52$$ and subtract. Subtracting each term individually from its similar power of x, we end up with $$0x^{3}+4x^{2}-16x+52$$. Since the $$0x^{3}$$ is the leading term and its value is $$0$$, it is insignificant, and we can get rid of it. We are now attempting to divide $$x^{2}-4x+13$$ into $$4x^{2}-16x+52$$. Again, we first look at the leading term on the outside and ask, what do we need to multiply it by to get the number on the inside? In this case, for $$x^{2}$$ to become $$4x^{2}$$, we multiply by a positive $$4$$. We write $$+4$$ on top of the division bar after the $$x$$ and then multiply the $$4$$ by the $$x^{2}-4x+13$$ to get $$4x^{2}-16x+52$$ which we subtract from the already written $$4x^{2}-16x+52$$ to get $$0$$. From here we can see that our long division is done. In particular, we have determined that $$x^{3}-3x+52$$ = $$(x^{2}-4x+13)(x+4)$$. But we already said that $$x^{3}-3x+52$$ = $$(x^{2}-4x+13)(x-c)$$, so $$c=-4$$. Thus, the final root of the polynomial $$x^{3}-3x+52$$ is $$x=-4$$, and the other two roots are $$x= 2+3i$$ and $$x=2-3i$$.

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