State the fundamental theorem of algebra (d'Alambert's theorem).
[answer]
Every non-constant single-variable polynomial \(a_nx^n + a_{n-1}x^{n-1} + \dots+a_0 =0\) with complex coefficients \(a_n, a_{n-1}, \dots, a_0\) has at least one complex root.
State the rational root theorem.
[answer]
Each rational root of \(a_nx^n + a_{n-1}x^{n-1} + \dots+a_0 =0\) is given by \(x= p/q\), where \(p\) is an integer factor of \(a_0\), and where \(q\) is an integer factor of \(a_n\).
Solve \(x^3 - x^2 - 5x -3 = 0.\)
[answer]
\(x=-1,-1,3\)
Solve \(x^3 - x^2 - 8 x - 6 = 0\).
[answer]
\(x=-1,1+\sqrt{7},1-\sqrt{7}\)
Solve \(x^4 - 2x^3 - 3x^2 + 8x -4 = 0\).
[answer]
\(x=-2,1,1,2\)
Decompose \(1/(x^2 + 2x-3)\) into partial fractions.
[answer]
Find the real part of \(e^{-ix}/(1+e^{a+ib})\).
[answer]
\begin{align*}
\frac{e^{-ix}}{1+ e^{a+ib}}&=\frac{\cos x - i\sin x}{1 + e^ae^{ib}}\\ &=\frac{\cos x -i\sin x}{1 + e^a(\cos b + i\sin b)}\\
&=\frac{\cos x -i\sin x}{1 + e^a(\cos b + i\sin b)}\frac{1 + e^a(\cos b - i\sin b)}{1 + e^a(\cos b - i\sin b)}\\
&= \frac{\cos x +e^a\cos x(\cos b - i\sin b)- i\sin x - ie^a\sin x(\cos b - i\sin b)}{1 +e^{a}(\cos b +i\sin b) + e^{a}(\cos b -i\sin b) + e^{2a}(\cos b +i\sin b)(\cos b -i\sin b)}\\
&= \frac{\cos x +e^a\cos x\cos b - ie^a\cos x\sin b- i\sin x - ie^a\sin x\cos b -e^a\sin x \sin b}{1 +e^{a}(\cos b) + e^{a}(\cos b) + e^{2a}(\cos^2 b + \sin^2 b)}\\
&=\frac{\cos x +e^a\cos x\cos b -e^a\sin x \sin b}{1 + e^{2a}(\cos^2 b + \sin^2 b)} -i\frac{e^a\cos x\sin b+\sin x +e^a\sin x\cos b}{1 + 2e^{a}\cos b + e^{2a}(\cos^2 b + \sin^2 b)}
\end{align*}
The real part is \(\frac{\cos x +e^a\cos x\cos b -e^a\sin x \sin b}{1 + e^{2a}(\cos^2 b + \sin^2 b)}\).
State the fundamental theorem of calculus.
[answer]
The derivative and integral are inverses.
What is the relationship between differentiability and continuity?
[answer]
Differentiability implies continuity, but continuity does not imply differentiability. For an example of a continuous function that is not differentiable, consider \(\sqrt{|x|}\) at \(x= 0\).
\(\lim_{x\to 0}\frac{e^x-1}{x^2+x}=\)
[answer]
\(1\) by L'Hopital's rule.
\(\lim_{x\to 0}\frac{2\sin x - \sin 2x}{x - \sin x}=\)
[answer]
\(6\) by repetitive use of L'Hopital's rule.
\(\lim_{x\to \infty} x^n\, e^{-x} =\)
[answer]
\(0\) by L'Hopital's rule. Repeat l'Hopital's rule until \(x^{n-1} = x^0\).
Show that \(\int u dv = uv - \int v du \).
[answer]
Start with the product rule, \(\frac{d}{dx} (uv) = v\frac{du}{dx} + u\frac{dv}{dx}\), integrate both sides over \(x\), and rearrange.
\(\int \ln x dx = \)
[answer]
\(x\ln (x) -x\)
\( \int e^x \sin x dx = \)
[answer]
Integrate by parts twice to get \( \int e^x \sin x dx = e^x[\sin(x)- \cos(x)]/2\).
What are some general guidelines about trigonometric subtitution?
[answer]
If a quantity \(a^2-x^2\) is involved, set \(x = a\sin\theta\). If a quantity \(x^2+ a^2\) is involved, set \(x = a\tan\theta\). If a quantity \(x^2-a^2\) is involved, set \(x = a\sec\theta\).
\( \int \sqrt{a^2-x^2} dx = \)
[answer]
\[\frac{a^2}{2}\arcsin{\frac{x}{a}} + \frac{a^2}{4}\frac{x}{a}[1-(x/a)^2]^{1/2} +C.\] See this page for more on trigonometric substitution.