Practise: Change, Slopes and Approximation

The fixed point solves $L=L/4+3$. Subtract $L/4$ from both sides: $3L/4=3$, so $L=4$. Now write the distance from the fixed point. Subtracting 4 from the rule gives $$x_{k+1}-4=\frac{x_k}{4}+3-4=\frac{x_k-4}{4}.$$ The starting distance is $1-4=-3$, so after $k$ steps the distance is $3/4^k$. For $k=4$ it is $3/256=0.01171875$, which is above $0.01$. For $k=5$ it is $3/1024=0.0029296875$, which is below it. Five steps are enough and four are not. The run reads $1$, $3.25$, $3.8125$, $3.953125$, $3.98828125$, $3.9970703125$, so the values creep toward 4 from below while the gap shrinks by the same factor at every step. The factor is what makes the guarantee: the distance is bounded by $3/4^k$ for every step past the fifth, not only at the fifth.

The top factors as $(x-4)(x+4)$. For $x$ different from 4 the factor $x-4$ is not zero, so it cancels: $$\frac{x^2-16}{x-4}=\frac{(x-4)(x+4)}{x-4}=x+4.$$ At $x=3.99$ this gives $3.99+4=7.99$, and at $x=4.01$ it gives $4.01+4=8.01$. The values on both sides sit close to 8 and close to each other, so the limit is 8. The original expression has no value at 4, and the limit does not need one: it records where the values are heading as the input approaches the hole. If you define the function to equal 8 at $x=4$, the new function is continuous there, because its limit equals its value. The cancellation is what supplies the missing value.

The second point is $5.1$, and $f(5.1)=5.1^2=26.01$. The rise is $26.01-25=1.01$, the run is $0.1$, and the quotient is $1.01/0.1=10.1$. The algebra shows where the number comes from: $$\frac{(5+h{)}^2-25}{h}=\frac{25+10h+h^2-25}{h}=\frac{10h+h^2}{h}=10+h.$$ At $h=0.1$ this is $10.1$. As $h$ tends to zero the leftover $h$ tends to zero, so the quotient tends to 10. The quotient is the slope of the secant through $(5,25)$ and $(5.1,26.01)$, and 10 is the slope of the tangent at $(5,25)$. For the square the derivative is $2x$, which equals 10 at $x=5$, so the limit and the formula agree. The gap is the whole difference between the two numbers: set it to $0.01$ and the quotient is $10.01$, and set it to $0.001$ and it is $10.001$.

The value is $f(2)=2^3=8$. The derivative is $f^{\prime }(x)=3x^2$, so the slope at 2 is $3\cdot 4=12$. The tangent line is $$T(x)=8+12(x-2)=12x-16.$$ At $x=2.01$ the gap is $0.01$, so the estimate is $8+12\cdot 0.01=8.12$. The true cube is $2.01^3=8.120601$, so the estimate is low by $0.000601$. The first correction came from the slope; the missed part comes from curvature, and its size is about half the second derivative times the square of the gap. Here $f^{\prime \prime }(x)=6x$, which is 12 at $x=2$, and $\frac{1}{2}\cdot 12\cdot (0.01{)}^2=0.0006$. The remaining $0.000001$ is the cubic term. The tangent line is the best straight-line stand-in at the point, and the cube leaves it because the cube bends.

Every derivative of $e^x$ equals $e^x$, so every derivative at 0 equals 1. The Taylor polynomial through degree three is $$T_3(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}.$$ At $x=0.1$ the terms are $1$, $0.1$, $0.005$ and $0.000166666\dots$. Their sum is $1.105166666\dots$, which rounds to $1.105167$ at six decimal places. The true value is $e^{0.1}=1.105170918\dots$, so the error is about $0.0000043$. The degree-two polynomial gave $1.105$ with an error of about $0.00017$, so the cubic term bought two more correct digits. Each extra term needs one more derivative at the expansion point and one more multiplication at evaluation, and the remainder tells you when to stop: keep terms until its bound falls below the tolerance you set.

The partial derivative in $x$ treats $y$ as a constant and gives $4x$, which is 4 at $x=1$. The partial derivative in $y$ treats $x$ as a constant and gives $2y$, which is 6 at $y=3$. The gradient is the list $$\nabla f(1,3)=(4,6).$$ Its length is $\sqrt{4^2+6^2}=\sqrt{52}\approx 7.211$, which is the rate of climb in the steepest direction. The downhill direction is $(-4,-6)$. A step of size $0.1$ in that direction subtracts $0.1(4,6)=(0.4,0.6)$ from the point, giving $(0.6,2.4)$. The height there is $$2(0.6{)}^2+(2.4{)}^2=0.72+5.76=6.48,$$ against $f(1,3)=2+9=11$ at the start. One step removed $4.52$ of height. The gradient at the point is perpendicular to the contour through it, so a step along the negative gradient crosses that contour at a right angle. That is the same statement as steepest descent, read on the map rather than on the slope.

Set $u=x^2+1$, so $y=u^3$. The inner rate is $du/dx=2x$, which is 4 at $x=2$. The outer rate is $dy/du=3u^2$, which is $3\cdot 25=75$ at $u=5$. The chain rule gives $$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}=75\cdot 4=300.$$ A direct check expands the cube: $(x^2+1{)}^3=x^6+3x^4+3x^2+1$, whose derivative is $6x^5+12x^3+6x$. At $x=2$ that is $192+96+12=300$. A small nudge of $0.001$ in $x$ raises the inner value by $0.004$, and the outer rate of 75 turns that into a rise of about $0.3$ in $y$. Divide by $0.001$ and the rate is about 300. The two local rates multiply because the nudge passes through both stages, and each stage scales it once.

The values are $1.51^4=5.19885601$, $1.5^4=5.0625$ and $1.49^4=4.92884401$. The forward quotient is $$\frac{5.19885601-5.0625}{0.01}=\frac{0.13635601}{0.01}=13.635601,$$ and the central quotient is $$\frac{5.19885601-4.92884401}{0.02}=\frac{0.270012}{0.02}=\mathrm{13.5006.}$$ The exact derivative is $4x^3=4\cdot 3.375=13.5$. The forward estimate is high by $0.135601$, and the central estimate is high by $0.0006$, so the central estimate is closer by a factor of about 226. The reason is the shape of the two errors: the forward quotient carries a term proportional to the step, while the central quotient cancels that term and carries one proportional to the step squared. Both numbers are estimates of the derivative at one point, and neither is a proof that the formula $4x^3$ is right everywhere. A second check at another point costs one more evaluation and tests a different part of the formula.