Chapter 0 wrap-up

Readiness: what numerical analysis leans on

Closing thought

These six tools answer one question from different sides: how close is a guess, and does the next guess get closer? Taylor's theorem gives an exact formula for the gap between a function and its polynomial, with one unknown point to bound. The definition of a limit says what "getting closer" means, and some sequences get there much faster than others. Elimination, eigenvalues and norms are exact linear algebra that later chapters will run with rounded numbers. Euler's method is the plainest case of stepping forward one guess at a time.

You can now put a number on how wrong an approximation could be. Take $\sin x\approx x$ at $x=0.1$ . Since $|\sin''|\le1$ , the remainder says the gap is at most $\frac{0.1^2}{2}=0.005$ . The true gap is about $0.00017$ , well inside that bound. You can also row-reduce a small system, find the eigenvalues of a $2\times2$ matrix, and measure a vector three ways. Next, Chapter 1 asks what happens when a computer stores every number with only so many digits. It builds the words and measures for how much error that causes.

Chapter review

One practice set per topic this chapter taught — same fixed, ungraded ladders as the practice panels inside each section.

Practice: Readiness check: Taylor remainders, limits, and Euler's method (6 problems · doesn’t affect your progress)

In the remainder $R_n(x)=\dfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$ , what is $\xi$ ?

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Why

Taylor's theorem promises a $\xi$ between $a$ and $x$ that makes the formula exact, without saying where it is. That is why we bound $f^{(n+1)}$ over the whole interval instead of evaluating it at one point.

Some unknown point strictly between $a$ and $x$ .

Approximate $\ln(1+x)$ by $x$ near $0$ . Here $f''(x)=-\frac{1}{(1+x)^2}$ . Using the remainder $R_1$ , what is the largest the gap can be at $x=0.1$ ?

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Why

$|R_1(0.1)|\le\frac{1}{2!}\cdot1\cdot0.1^2=0.005$ . The true gap, $0.1-\ln1.1\approx0.0047$ , sits just inside the bound.

Correct answer

Also accepted:

Why

On $[0,0.1]$ , the biggest value of $\frac{1}{(1+\xi)^2}$ is at $\xi=0$ .

0.005

$x_n=\frac{2n+3}{n}$ converges to $2$ . What is the smallest $n$ with $|x_n-2|<0.001$ ?

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Why

$|x_n-2|=\frac3n$ , and $\frac3n<0.001$ needs $n>3000$ . So $n=3001$ is the first index that works.

Correct answer

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Why

3001

Both $\frac1n$ and $\left(\frac12\right)^n$ go to $0$ . Which one first gets within $10^{-6}$ of $0$ , and roughly when?

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Why

$2^{20}\approx1.05\times10^6$ , so $\left(\frac12\right)^{20}<10^{-6}$ . For $\frac1n<10^{-6}$ you need $n>10^6$ . Same limit, very different speed. Never reaching $0$ exactly doesn't stop either sequence from converging.

$\left(\frac12\right)^n$ , at about $n=20$ .

For $y'=2y$ , $y(0)=1$ , run Euler's method with $h=0.25$ for two steps. What is the estimate of $y(0.5)$ ?

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Why

$y_1=1+0.25(2)=1.5$ , then $y_2=1.5+0.25(3)=2.25$ . The true value is $e^{1}\approx2.718$ , so Euler's straight-line steps fall short.

Correct answer

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Why

2.25

For $y'=y$ , $y(0)=1$ , run Euler's method to $t=1$ with $h=0.25$ (four steps). What is $y_4$ ? Round to 4 decimal places.

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Why

Each step multiplies by $1.25$ , so $y_4=1.25^4\approx2.4414$ . With $h=0.5$ you'd get $1.5^2=2.25$ . The true $e\approx2.7183$ is closer to the smaller step's answer.

Correct answer

Also accepted:

Why

Each step multiplies the current value by $1+h$ .

2.4414

Practice: Readiness check: elimination, eigenvalues, and norms (6 problems · doesn’t affect your progress)

Solve $x+2y=5$ , $3x+4y=6$ by Gaussian elimination. What is $y$ ?

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Why

Subtract $3$ times row 1 from row 2: $-2y=-9$ , so $y=4.5$ . Back-substitution then gives $x=5-9=-4$ .

Correct answer

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Why

4.5

A system's first pivot is $0.0001$ . Eliminating below it means subtracting $10{,}000$ times row 1 from row 2. Why can that hurt when the arithmetic is floating point?

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Why

In exact arithmetic any nonzero pivot works. With a fixed number of digits, adding $10{,}000\times$ row 1 makes row 2's original entries tiny by comparison, and rounding throws their digits away. Swapping rows first avoids the tiny pivot.

Row 2's entries get swamped, and rounding drops their digits.

$A=\begin{pmatrix}4&1\\2&3\end{pmatrix}$ . What is its larger eigenvalue?

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Why

$\det(A-\lambda I)=(4-\lambda)(3-\lambda)-2=\lambda^2-7\lambda+10=(\lambda-5)(\lambda-2)$ . The eigenvalues are $5$ and $2$ .

Correct answer

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Why

The characteristic polynomial is $\lambda^2-7\lambda+10$ .

What are the eigenvalues of $\begin{pmatrix}2&5\\0&-3\end{pmatrix}$ ?

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Why

The matrix is triangular, so $\det(A-\lambda I)=(2-\lambda)(-3-\lambda)$ . The eigenvalues are the diagonal entries. The $5$ above the diagonal doesn't enter. The pair $-1$ and $-6$ is the trace and determinant, not the eigenvalues.

$2$ and $-3$

What is the $\ell_2$ norm of $\mathbf v=(3,-4,12)$ ?

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Why

$\sqrt{9+16+144}=\sqrt{169}=13$ .

Correct answer

Also accepted:

Why

For $\mathbf v=(3,-4,12)$ , which ordering of its three norms is correct, and does it hold for every vector?

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Why

Compute all three: the largest entry, the length, and the sum of absolute values.

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