Abstract
Quiz 1 in the course Infinitesimal Calculus (2), semester 2026 B.
This is a quiz that I translated and solved.
The original quiz was written by the Infi Calculus 2 course lecturers at HUJI.
Problem 1. Find a \(q\in\mathbb{Q}\text{ s.t.}\left|e^{-1}-q\right|<10^{-2}\).
Solution. Our goal is to find a rational approximation for the function \(e^{x}\) evaluated at \(x=-1\), with an error strictly less than \(10^{-2}\).
The closest point to \(x=-1\) which we can easily evaluate is \(x=0\) since \(e^{0}=1\).
The Maclaurin polynomial (the Taylor polynomial at 0) of degree \(n\) of \(e^{x}\) is \[P_{n,e^{x},0}(x)=\underset{k=0}{\overset{n}{\sum}}\frac{f^{(k)}(0)}{k!}x^{k}=e^{0}+e^{0}\cdot x+\frac{e^{0}}{2!}\cdot x^{2}+\dots+\frac{e^{0}}{n!}\cdot x^{n}=1+x+\frac{x^{2}}{2!}+\dots+\frac{x^{n}}{n!}\]
Let \(n\in\mathbb{N}\). To apply Taylor’s Theorem on \(e^{x}\) in the interval \(\left[-1,0\right]\), \(e^{x}\) must have a continuous \(n\)-th derivative on \(\left[-1,0\right]\) and be differentiable \(n+1\) times on \(\left(-1,0\right)\).
\(e^{x}\) is differentiable \(n\) times on \(\left[-1,0\right]\) with its \(n\)-th derivative being \(e^{x}\), which is continuous on \(\mathbb{R}\) and in particular on \(\left[-1,0\right]\).
\(e^{x}\) is also differentiable \(n+1\) times on \(\left(-1,0\right)\).
In fact, \(e^{x}\) is a smooth function (it is infinitely differentiable) on \(\mathbb{R}\).
Hence, the Taylor Theorem with the Lagrange Remainder applies and there exists a point \(c\in\left(-1,0\right)\) such that \[R_{n,e^{x},0}(-1)=e^{-1}-P_{n,e^{x},0}(-1)=\frac{e^{c}}{\left(n+1\right)!}\left(-1\right)^{n+1}\] This means that \[\left|e^{-1}-P_{n,e^{x},0}(-1)\right|\le\left|\frac{e^{c}}{\left(n+1\right)!}\left(-1\right)^{n+1}\right|=\left|\frac{e^{c}}{\left(n+1\right)!}\right|\underset{\left(*\right)}{\le}\frac{1}{\left(n+1\right)!}\]
\(\left(*\right)\) since \(e^{x}\) is a monotonically increasing function, \(\forall x\in\left(-1,0\right),e^{x}\le e^{0}=1\).
To guarantee our error is strictly less than \(10^{-2}\), we need \(\frac{1}{(n+1)!}<\frac{1}{100}\), which implies \((n+1)!>100\).
Since \(4!=24\) and \(5!=120\), the minimum degree required is \(n=4\).
Therefore, setting the degree \(n=4\) gives the rational approximation: \[P_{4,e^{x},0}\left(-1\right)=1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}=\frac{3}{8}\in\mathbb{Q}\]
which satisfies the error bound: \[\left|e^{-1}-P_{4,e^{x},0}(-1)\right|\le\frac{1}{5!}<\frac{1}{100}=10^{-2}\]
as required.
Problem 2. Find all antiderivatives of \(f(x)=\frac{x^{2}-4x}{x^{2}-3x+2}\) on \(\left(1,2\right)\).
Solution. Since the degree of the numerator is greater than or equal to the degree of the denominator, our first step is to perform polynomial division with a remainder.
We can achieve this by expressing the numerator in terms of the denominator. \[f(x)=\frac{x^{2}-3x+2-x-2}{x^{2}-3x+2}=1+\frac{-x-2}{x^{2}-3x+2}\]
Next, we focus on finding the antiderivatives of the remainder, \(g(x)=\frac{-x-2}{x^{2}-3x+2}\), on the interval \((1,2)\).
As a best practice, we first check if the numerator is a scalar multiple of the denominator’s derivative, which would allow for direct logarithmic integration. \[\left(x^{2}-3x+2\right)'=2x-3\]
Since this is not the case, we proceed with partial fraction decomposition.
We first split the denominator into coprime polynomials. \[x^{2}-3x+2=\left(x-2\right)\left(x-1\right)\]
Hence \[g(x)=\frac{-x-2}{\left(x-2\right)\left(x-1\right)}\] \[\frac{-x-2}{\left(x-2\right)\left(x-1\right)}=\frac{A}{x-2}+\frac{B}{x-1}\] \[-x-2=\left(x-1\right)A+\left(x-2\right)B\] \[\left(A+B\right)x-A-2B=-x-2\]
And since two polynomials are equal if and only if their corresponding coefficients are equal, \[\begin{cases} A+B=-1\\ -A-2B=-2 \end{cases}\Rightarrow\begin{cases} A=-4\\ B=3 \end{cases}\] \[g(x)=\frac{-4}{x-2}+\frac{3}{x-1}\] Now we can return to our original function, \(f(x)=1+g(x)\), and find the antiderivatives term by term.
The antiderivative of 1 is \(F_{1}(x)=x+C\).
The antiderivative of \(\frac{-4}{x-2}\) is \[F_{2}(x)=-4\ln\left|x-2\right|+C\] and the antiderivative of \(\frac{3}{x-1}\) is \[F_{3}(x)=3\ln\left|x-1\right|+C\]
Therefore, the general antiderivative of the function is \[F(x)=x-4\ln\left|x-2\right|+3\ln\left|x-1\right|+C\]
For \(x\in(1,2)\), we have \(x-2<0\) and \(x-1>0\). Thus, the antiderivative simplifies to: \[F(x)=x-4\ln\left(2-x\right)+3\ln\left(x-1\right)+C\]