5 Answers
5.1 Powers and logarithms
Simplify the following expressions
Exercise 1.
\(5^{1/3} \cdot 200^{1/3} = 5^{1/3} \cdot 2^{1/3} \cdot 100^{1/3} = 10^{1/3} \cdot 100^{1/3} = 10^{1/3} \cdot 10^{2/3} = 10^1 = 10\)
Exercise 2.
\(\sqrt{1+x}(1+x)^{3/2}= (1+x)^{1/2}(1+x)^{3/2} = (1+x)^{1/2+3/2} = (1+x)^2\)
Exercise 3.
\(\dfrac{a^{x-1}}{a^{x+2}} = a^{x - 1 - (x + 2)} = a^{-3} = \dfrac{1}{a^3}\)
Exercise 4.
\(\dfrac{1}{\sqrt{e^{-x}e^{7x}}} = \dfrac{1}{\sqrt{e^{6x}}} = {(e^{6x})}^{\frac{-1}2}=e^{-3x}\)
Exercise 5.
\(\dfrac{\ln(a)}{\ln(b^x)+\ln(c^x)} e^{\ln(e^5)} = \dfrac{\ln a}{x\ln b+x\ln c} e^5 = \dfrac{\ln a}{x\ln(b.c)} e^5 = \dfrac{e^5}{x}\log_{bc} (a)\)
Exercise 6.
\(\ln{\left(x^2-2x+1\right) - \ln{\left(x - 1\right)}} = \ln{\left( {(x-1)}^2 \right)} - \ln{(x - 1)} = 2\ln{(x-1)} -\ln{(x-1)} = \ln{(x-1)}\)
Solve the following expressions for \(x\) using factorization where possible
Exercise 7.
\(x^2-\sqrt{2}x-\frac{5}4 = 0\)
\(A = 1, B=-\sqrt{2}, C=-\frac{5}4\) and \(x = \frac{-B \pm \sqrt{B^2-4AC}}{2A}\)
\(x = \frac{\sqrt{2} \pm \sqrt{2-(-5)}}{2}=\frac{\sqrt{2}\pm\sqrt{7}}2\)
Exercise 8.
\(9x^2-12x+4= 0\)
\(A =9, B=-12, C=4\) and \(x = \frac{-B \pm \sqrt{B^2-4AC}}{2A}\)
\(x = \frac{12 \pm \sqrt{144-144}}{18}=\frac{12}{18} = \frac{2}3\)
Exercise 9.
\(5x^2-4x-1 = 0 \rightarrow (5x+1)(x-1)=0\)
\(x=1\) or \(x=-\frac{1}5\)
Exercise 10.
\(x^2 + 3 x - 10 = (x + 5)(x - 2)\), hence \(x = -5\) or \(x = 2\)
Solve the following expressions for \(x\)
Exercise 11.
\(\sqrt{{(2^3)}^x} = 64 \rightarrow {(2^{3x})}^{1/2} = 2^6\)
\(2^{\frac{3x}{2}} = 2^6 \rightarrow \frac{3x}{2} = 6\) then \(x = 12/3 = 4\)
Exercise 12.
\(e^{x^2-2x}=e^8 \rightarrow x^2-2x-8=0\)
\((x-4)(x+2)=0 \rightarrow x=4\) or \(x=-2\)
Exercise 13.
\(\ln(4-x) = \frac{1}2\)
Take the exponential power of both sides: \(e^{\ln(4-x) }= e^{\frac{1}2}\)
\((4-x) = e^\frac{1}2\rightarrow x =4-e^\frac{1}2 \simeq 2.351278729\)
Exercise 14.
\(\log_2 \left(5\right) + \log_2 \left(x\right)=\log_2 \left( 10 \right)\)
\(\log_2 \left(5x\right) =\log_2 \left( 10 \right)\rightarrow 5x = 10\), then \(x=2\).
Exercise 15.
\(\ln(\sqrt{x})=\sqrt{\ln x}\)
\(\ln(x^{\frac{1}2}) = {(\ln x)}^{\frac{1}2} \rightarrow \frac{1}2 \ln x = {(\ln x)}^{\frac{1}2}\)
We might take the square of both sides to get rid of the square root:
\(\frac{1}{4} {(\ln x)}^2 = \ln x\)
Now let’s give \(\ln x\) another name. Say \(y = \ln x\). Then we have: \(\frac{1}{4} y^2 = y\)
\(\frac{1}{4} y^2 - y = 0 \rightarrow y(\frac{1}{4} y - 1) = 0\)
Then \(y=0\) or \(\frac{1}{4} y - 1 = 0 \rightarrow y = 4\)
Remember that \(y\) was actually \(\ln x\), and we were trying to calculate \(x\).
So: \(\ln x = 0 \rightarrow x = 1\) or \(\ln x = 4 \rightarrow x = e^4 \approx 54.5982\)
Exercise 16.
\(5 \ln x - 2 \ln x^2 = 5 \ln x - 4 \ln x = \ln x = 4\), hence \(x = e^4 \approx 54.6\)
Exercise 17.
\({\left( \dfrac{ x^2}{\sqrt{x}} \right)}^2 = \dfrac{x^4}{x} = x^3 = 125 \longrightarrow x = \sqrt[3]{125} = 5\)
Exercise 18.
\(\dfrac{a \ln{\left( {(x - 1)}^2 \right)}}{\ln{\left( {(x - 1)}^b \right)}} = \dfrac{2 a \ln{(x-1)}}{b\ln{(x-1)}} = \frac{2a}{b}\)
Hence, \(2a=b\). They should not be equal to \(0\).
Applications
Exercise 19. Free energy of a reaction
a. We can separate an expression with \([A]\) and \([B]\) from one with \(K_{eq}\): \[ \Delta G = R T \ln{ \left( \frac{[B]/[A]}{K_{eq}} \right)} = R T \ln{([B]/[A])} - R T \ln{(K_{eq})} \] in which \(- R T \ln{(K_{eq})}\) equals the standard Gibbs free energy \(\Delta G^0\)
Exercise 20. Information theory
a. \[ H = - \left( 0.2\log_{2}(0.2) + 0.25\log_{2}(0.25) + 0.35\log_{2}(0.35) + 0.2\log_{2}(0.2) \right) \approx 1.96 \] So, almost 2 bits.
b. \[ H = - \left( 0.1\log_{2}(0.1) + 0.1\log_{2}(0.1) + 0.4\log_{2}(0.4) + 0.4\log_{2}(0.4) \right) \approx 1.72 \] We gain less information because there is less information to gain from a sequence where we could predict in advance that certain symbols (G and C) occur much more often than others.
c. We gain maximal information (2 bits) at equal probabilities (0.25) for each of the symbols, and no information at all (0 bits) when one of the symbols has probabilty 1 of occurring.
Exercise 21. The genetic code
a. The total number of possible codons can be calculated using the formula \(4^3=64\) since a codon has three positions and each can be occupied by one of four nucleotides.
b. A minimal genetic code requires at least 21 codons: 20 for encoding amino acids and 1 for a stop signal. A system using 2-nucleotide codons would provide only \(4^2=16\) codons, which is insufficient to encode all necessary amino acids and a stop signal.
c. The function \(C(n)=4^n\) describes the number of possible codons, where \(n\) is the number of nucleotides per codon. This is an exponential function.
Exercise 22. The polymerase chain reaction
a. The number of DNA fragments after 15 cycles is \(N = 50 \times 2^{15}\). The power of \(2\) with an exponent of 15 is \(2^{15} = 32768\). Therefore \(N = 50 \times 32768\). Thus, the number of DNA fragments after 15 cycles is \(1638400\).
b. \[ \begin{align*} 10000000 &= 100 \times 2^n \\ 100000 &= 2^n \\ \log_2(100000) &= n \\ n &= \frac{\log_{10}(100000)}{\log_{10}(2)} \\ n &= \frac{5}{0.3010} \approx 16.61 \end{align*} \]
Since the number of cycles must be an integer, at least 17 cycles are required to produce at least 10 million DNA fragments from an initial number of 100.
c. \[ \begin{align*} 0.01 &= 0.1 e^{-5k} \\ 0.1 &= e^{-5k} \\ \ln(0.1) &= -5k \\ -2.3026 &= -5k \\ k &= 2.30165 \approx 0.4605 \end{align*} \]
Thus, the constant \(k\) is approximately \(0.4605\)
Exercise 23. DNA substitution
a. \[ \begin{align*} S(i,j) &= -2 \\ -2 &= \log_2 \left( \frac{p_{i,j}}{q_i \, q_j} \right) \\ \frac{p_{i,j}}{q_i \, q_j} &= 2^{-2} = \frac{1}{4} \end{align*} \] This means that the substitution from \(i\) to \(j\) is less likely than expected by a factor of 4.
b. The base-2 logarithm is often used in bioinformatics because it directly relates to binary comparisons. This makes the interpretation of log-odds ratios intuitive, as each unit increase corresponds to a doubling in likelihood. Using a different base, such as the natural logarithm, does not change their relative magnitudes, but it can make this interpretation less straightforward.
Exercise 24. Gene expression analysis
a. \[ \begin{align*} \log_2 \left(\text{FC}\right) &= \log_2 \left(\frac{500}{125}\right) \\ \log_2 \left(\text{FC}\right) &= \log_2(4) \\ \log_2 \left(\text{FC}\right) &= 2 \end{align*} \]
This means that this gene is more expressed in the Treatment group compared to the Control group.
b. \[ \begin{align*} \log_2 \left(\text{FC}\right) &= \log_2 \left( \frac{\text{Expression(Disease)}}{\text{Expression(Control)}} \right) = -2 \\ \frac{\text{Expression(Diseased)}}{\text{Expression(Control)}} &= 2^{-2} = \frac{1}{4} \end{align*} \]
This means that the expression of this gene in the Diseased group is \(\frac{1}{4}\) (or 4 times lower) of that in the Control group. This gene is downregulated in the Diseased condition by a factor of 4.
Exercise 25. Unit conversions
a. \(1\unit{m} = 100 \unit{cm}\), \(1\unit{m}^2 = 10^4 \unit{cm}^2\), and \(1\unit{m}^3 = 10^6 \unit{cm}^3\)
b. \(1 \unit{dm}^2 = 1 \unit{L}\), \(1 \unit{cm}^2 = 10^{-3}\unit{L}\), and \(1 \unit{nm}^3 = 10^{-24} \unit{L}\)
c. \(1 \unit{fL} = 10^{-15} \unit{L} \times 10^{6} \frac{\unit{\mu L}}{\unit{L}} = 10^{-9} \unit{\mu L}\), \(1 \unit{mL} = 10^{-3} \text{L}\), and \(1 \unit{\mu L} = 10^{-6} \unit{L} \times 10^{15} \frac{\unit{fL}}{\unit{L}} = 10^{9} \unit{fL}\)
d. \(1 \unit{mol} = 6\tothe{23} \unit{molecules}\), \(1 \unit{\mu mol} = 10^{-6} \unit{mol} \times 6\tothe{23} \frac{\unit{molecules}}{\unit{mol}} = 6 \tothe{17} \unit{molecules}\), and \(1 \unit{pmol} = 6 \tothe{11} \unit{molecules}\)
e. \(50 \unit{nM}\unit{min}^{-1} = 50 \unit{nM}\unit{min}^{-1} \times 10^{-9} \unit{M} \unit{nM}^{-1} \times 0.0167 \unit{min}\unit{s}^{-1} = 8.3 \tothe{-10} \unit{M}\unit{s}^{-1}\)
f. \(10^{-12} \unit{cm}^2\unit{s}^{-1} = 6 \tothe{-15} \unit{m}^2\unit{min}^{-1}\)
g. \(10^{16} \unit{cm}^{-3} = 1.67 \tothe{-5}\unit{M}\)
Exercise 26. The rate of the DNA polymerase
The time it takes equals \(4558953 \unit{bp} \times \frac{1}{800} \frac{\unit{s}}{\unit{bp}} \times 0.5 = \frac{4558953}{1600} \unit{s} = 2849 \unit{s} = 2849 \unit{s} \times \frac{1}{60} \frac{\unit{min}}{\unit{s}} = 47 \unit{min}\).
Exercise 27. Membrane proteins
a. We assume E. coli to be spherical. Then it has as area \(A_\text{cell} = 4 \times \pi \times r_\text{cell}^2\) with \(r_\text{cell}\) as the cell radius. Only 50% of this area can be occupied by protein, which is \(2 \times \pi \times r_\text{cell}^2\). The area of a disk equals \(A_\text{protein} = \pi \times r_\text{protein}^2\). So the number of proteins in the membrane of E. coli equals maximally; \[ \begin{align*} \frac{2 \times \pi \times r_\text{cell}^2 \frac{\unit{m}^2}{\unit{cell~membrane}}} {\pi \times r_\text{protein}^2 \frac{\unit{m}^2}{\unit{protein}}} = \\ \frac{2 \times \pi \times (10^{-6})^2 \unit{m}^2}{\pi \times (5\tothe{-9})^2 \unit{m}^2} \frac{\unit{protein}}{\unit{cell membrane}} = \\ \frac{2}{25} \times 10^{6} \unit{\frac{protein}{\text{cell membrane}}} = \\ 80000 \frac{\unit{protein}}{\unit{cell membrane}} \end{align*} \]
b. Next, we calculate the volume of the periplasm of an cell using the equation of the volume of a sphere \(V = \frac{4}{3}\times \pi \times r^3\): \(V_\text{periplasm} = \frac{4}{3} \times \pi \times \left(10^{-6} \unit{m} + 15\tothe{-9} \unit{m} \right)^3 - \frac{4}{3} \times \pi \times \left(10^{-6} \unit{m} \right)^3 = 1.9\tothe{-19} \unit{m}^3\). The volume of a protein equals \(V_\text{protein} = \frac{4}{3} \times \pi \times (5\tothe{-9} \unit{m})^3 = 5.2\tothe{-25} \unit{m}^3\). So, maximally \(\frac{1.9\tothe{-19}}{5.2\tothe{-25}} = 0.4\) million protein molecules fit in the periplasmic space. Which is an appreciable amount of the entire protein content of a cell. All those proteins need to be transported over the plasma membrane during cell growth! This requires an enormous transport capacity of a cell especially at maximal growth rates.
c. We calculate the volume of the cytoplasm: \(V_\text{cytoplasm} = \frac{4}{3} \times \pi \times \left(10^{-6} \unit{m}\right)^3\), the volume of proteins is the same, so again a maximum of \(\frac{4.2\tothe{-18}}{5.2\tothe{-25}} = 8\tothe{6}\) proteins fit in the cytoplasm. The ratio then becomes \(\frac{8\tothe{4} + 4 \times 10^5}{8 \tothe{6}}=\frac{4.8 \tothe{5}}{80 \tothe{5}} = 0.06\) or 6%.
5.2 Functions
Sketch these functions
Exercise 1.
\(y = - e^x\rightarrow\) Mirror image of \(e^x\) with respect to \(x\)-axis.
Exercise 2.
\(y=- \ln x \rightarrow\) Mirror image of \(\ln x\) with respect to \(x\)-axis.
Exercise 3.
\(y = 2^{x-3}\)
\(y = \frac{2^x}{8} \rightarrow\) So it means the graph will scale down by a factor 8.
Exercise 4.
\(g(x) = |x-2|\)
Analyze these functions
Exercise 5.
- \(f(0) = -\frac{1}{2}\) and \(f(x) = 0\) for \(x = -1\).
- \[ \begin{align*} f'(x) & = \frac{(x-2) \cdot 1 - (x+1) \cdot 1}{{(x-2)}^2} \\ & = \frac{-3}{{(x-2)}^2} \end{align*} \]
This is never equal to 0, no minima or maxima.
- \(f''(x) = \frac{6}{{(x-2)}^3}\) is never 0.
- \(\lim_{x \to 2}f(x) = \infty\)
- \(\lim_{x \to \infty}f(x) = 1\) and \(\lim_{x \to -\infty}f(x) = 1\)
Exercise 6.
- Intersections with the \(x\)-axis (\(f(x)=0\)): \(x \ln{x} = 0\). \(x=0\) is not a solution, because \(\ln{0}\) is not defined. When \(\ln{x}=0\) then \(x = e^0 = 1\). Hence, \((1,0)\) is an intersection point.
- \[ \begin{align*} f'(x) & = x \cdot \frac{1}{x} + 1 \cdot \ln x \\ & = 1 + \ln x \end{align*} \]
This is 0 for \(x = \frac{1}{e}\). \(f(\frac{1}{e}) = -\frac{1}{e}\).
\(f''(x) = \frac{1}{x}\), thus \(f''(\frac{1}{e})\) is positive and this is a minimum. \(f''(x)\) is never 0 \(\lim_{x \to 0}x \ln{x} = 0\), (Use L’ Hospital’s rule on \(\dfrac{\ln{x}}{1/x}\), or just try decreasing \(x\) to \(0\) on your calculator). There is no vertical asymptote, i.e. \(x \ln{x}\) is defined everywhere for \(x>0\).
Exercise 7.
Limits
Exercise 8.
\(\lim_{x \to 3} \dfrac{x^2 - 2x - 3}{x^2 - 9} = \lim_{x \to 3} \dfrac{(x - 3)(x + 1)}{(x - 3)(x + 3)} = \lim_{x \to 3} \dfrac{x + 1}{x + 3} = \frac{4}{6}\)
Exercise 9.
\(\lim_{x \to \infty} \dfrac{ax}{b + x} = \lim_{x \to \infty} \dfrac{a}{\left(\frac{b}{x}\right) + 1}\). The quantity \(\frac{1}{x}\) tends to \(0\) as \(x \to \infty\). The remainder of the expression is well determined. Hence: \(\lim_{x \to \infty} \dfrac{ax}{b + x} = a\)
Exercise 10.
\(\lim_{x \to \infty} \dfrac{ax}{b^2 + x^2} = \lim_{x \to \infty} \dfrac{a}{\left(\frac{b^2}{x}\right) + x} = \lim_{x \to \infty}\). The quantity \(\frac{b^2}{x}\) tends to \(0\) as \(x \to \infty\). Hence, the denominator increases to \(\infty\) as \(x \to \infty\). Hence: \(\lim_{x \to \infty} \dfrac{ax}{b^2 + x^2} = 0\)
Applications
Exercise 11.
a. \(a>0\) thus we have to show that \(e^{-bt} - e^{-ct} > 0\):\ \(e^{-bt} - e^{-ct} = \frac{1}{e^{bt}} - \frac{1}{e^{ct}}\). Because \(t >0\) and \(c>b\): \(\frac{1}{e^{bt}} > \frac{1}{e^{ct}}\) and \(\frac{1}{e^{bt}} - \frac{1}{e^{ct}} >0\)
b. \[ \begin{align*} \frac{dy}{dt} &= a(-b e^{-bt} + c e^{-ct}) = 0 \\ t &= \frac{1}{b-c} \ln \frac{b}{c} \\ y \left( \frac{1}{b-c} \ln \frac{b}{c} \right) &= a \left( \frac{b}{c}^{-\frac{b}{b-c}}-\frac{b}{c}^{-\frac{c}{b-c}} \right) \end{align*} \]
c. \[ \begin{align*} \frac{d^2y}{dt^2} &= a(-b^2 e^{-bt} + c^2 e^{-ct}) = 0 \\ t &= \frac{1}{b-c} \ln \frac{b^2}{c^2} \end{align*} \]
d.
Exercise 12. Time complexity of algorithms
a. \(T_A(n) = n^2\) will be a parabola opening upwards, while \(T_B(n) = 50n\) will be a straight line with a slope of 50.
b. - \(T_A(30) = 30^2 = 900\) and \(T_B(30) = 50 \times 30= 1500\). Thus, Algorithm A is faster for \(n=30\). - \(T_A(80) = 80^2 = 6400\) and \(T_B(80) = 50 \times 80= 4000\). Thus, Algorithm B is faster for \(n=80\).
c. \[ \begin{align*} n^2 &= 50 n \\ n^2 - 50n &= 0 \\ n (n − 50) &= 0 \end{align*} \]
The graphs intersect at \(n = 0\) and \(n = 50\). Hence, the algorithms are equally fast for sequences of length 50. For shorter sequences, algorithm A is faster. For longer sequences, algorithm B is faster.
d. It is \(n^2\) for A and \(n\) for B.
Exercise 13. Comparing efficiency of two algorithms
\(O(C_1(n))\) is \(n^2\) and \(O(C_1(n))\) is \(n \log(n)\), because these are the dominant terms in these functions. The term \(n^2\) dominates \(n\log(n)\), therefore A2 is more effcicient for large \(n\).
Exercise 14. Activation functions for neural networks
a. Domains are \(\mathbb{R}\) in all four cases. Ranges or images are 1: \(\{0,1\}\); 2: \((0,1)\); 3: \([0,\infty)\); 4: \((-\infty,\infty)\).
b. 1. \(\dd{\mathcal{H}(x)}{x} = 0\) for all values of \(x\), except when \(x=0\): in that case it is undefined. 2. \[ \dd{\phi(x)}{x} = \frac{e^{-x}}{\left( 1 + e^{-x}\right)^2} \] 3. \[ \dd{R(x)}{x} = \begin{cases} x < 0 & 0 \\ x \geq 0 & 1 \end{cases} \] 4. \[ \dd{LR(x)}{x} = \begin{cases} x < 0 & 0.1 \\ x \geq 0 & 1 \end{cases} \]
The disadvantage of the Heaviside function is that its derivative is equal to 0 everywhere, and undefined at \(x=0\). An algorithm using an update proportional to the derivative will fail with this function. The disadvantage of the logistic function is that its derivative is nearly \(0\) at high and low positive values of \(x\). An algorithm that updates weights in proportion to the derivative may fail or adapt very slowly here.
5.3 Differentiation
Differentiate the following functions
Exercise 1.
One of the “elementary” derivatives \(f'(x)=\dfrac{1}{x}\)
Exercise 2.
One of the “elementary” derivatives: \(f'(x) = nx^{n-1}\)
Exercise 3.
Since \(f(x) = \frac{1}{\sqrt[5]{x}} = \dfrac{1}{x^{\frac{1}{5}}} = x^{\frac{-1}{5}}\) we have an elementary function of the form \(f(x)=x^n\). Hence, \(f'(x)=-\frac{1}{5}x^{-\frac{6}{5}}=-\frac{1}{5\sqrt[5]{x^{6}}}=-\frac{1}{5x\sqrt[5]{x}}\)
Exercise 4.
\(f(x) = \dfrac{1}{\sqrt[3]{x}} = x^{-1/3}\). Then \(f'(x) = \dfrac{-1}{3} x^{-4/3} = \dfrac{-1}{3 x \sqrt[3]{x}}\)
Exercise 5.
\(y'=6x-5\)
Exercise 6.
\(f'(x) = 3x^{2}-x^{-2}\)
\(f''(x) = 6x+\frac{2}{x^{3}}\)
Exercise 7.
\(f'(x) = x \cdot \dfrac{1}{x} + \ln x \cdot 1 = 1 + \ln x\)
Exercise 8.
\(f'(x)=6x\cdot(e^{5x}-x^{5})+(3x^{2}+5)(5e^{5x}-5x^{4})\)
Exercise 9.
\(f'(x) = 2 \cdot 10^6 x\)
Exercise 10.
\(f'(x) = \dfrac{ad-bc}{{(cx+d)}^2}\)
Exercise 11.
Met behulp van de quotientregel
\[ f'(t) = \frac{-3 \cdot t^{-2}}{t^6} = \frac{-3}{t^{4}} \] of door te herschrijven \(f(t) = t^{-3}\):
\[ f(t) = -3 \cdot t^{-4} = \frac{-3}{t^{4}} \]
Exercise 12.
\(f'(x)=\frac{1}{2}{(25-x^{2})}^{\frac{-1}{2}} \cdot (-2x)=\frac{-x}{\sqrt{(25-x^{2})}}\)
Exercise 13.
\(f'(x) = \frac{1}{2}{(1 + e^{3x})}^{-\frac{1}{2}} \cdot e^{3x} \cdot 3 = \dfrac{3 e^{3x}}{2 \sqrt{1 + e^{3x}}}\)
Exercise 14.
\(f'(x) = \dfrac{1}{x + 1}\)
Exercise 15.
\(f'(x) = -\frac{1}{2} e^{-x^2} \cdot -2 x = x e^{-x^2}\)
Exercise 16.
\(f'(x)=2x\cdot e^{x^{2}}-49e^{7x}\)
Exercise 17.
\(f'(x) = \ln{(x + 1)} + (x + 1) \frac{1}{x + 1} = \ln{(x + 1)} + 1\)
Exercise 18.
\(f'(x) = 3{(x^{2} - \ln x)}^2 \cdot (2x - \frac{1}{x}\))
Exercise 19.
\(f'(x) = \dfrac{n x^{n-1}(1 + x^n) - n x^{n-1} x^n}{{(1 + x^n)}^2} = \dfrac{n x^{n-1}}{{(1+x^n)}^2}\)
Exercise 20.
First simplify: \(f(x) = \ln{\left( \dfrac{x(x+1)(x+5)}{x(x+5)} \right)} = \ln{(x + 1)}\). Then applying the chain rule \(f'(x) = 1 \cdot \dfrac{1}{x+1} = \dfrac{1}{x+1}\)
Differentiation from first principles
Exercise 21.
\[ \begin{align*} \dd{y}{x} & = & \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{{(x + h)}^3 - x^3}{h} \\ &= \lim_{h \to 0} \frac{(x + h)(x^2 + 2xh + h^2) - x^3}{h} \\ &= \lim_{h \to 0} \frac{x^3 + 3x^2 h + 3xh^2 + h^3 - x^3}{h} \\ &= \lim_{h \to 0} \frac{3x^2 h + 3x h^2 + h^3}{h} \\ &= \lim_{h \to 0} 3x^2 + 3xh + h^2 \\ &= 3x^2 \end{align*} \]
Exercise 22.
\[ \begin{align*} \dd{f}{x} &= \lim_{h \to 0}\frac{\frac{1}{{(x + h)}^2} - \frac{1}{x^2}}{h} \\ &= \lim_{h \to 0} \frac{\frac{x^2 - {(x + h)}^2}{x^2 {(x + h)}^2}}{h} \\ &= \lim_{h \to 0} \frac{x^2 - (x^2 + 2xh + h^2)}{x^2 h {(x + h)}^2} \\ &= \lim_{h \to 0} \frac{-2xh -h^2}{x^2 h {(x + h)}^2} \\ &= \lim_{h \to 0} \frac{-2x - h}{x^2 {(x + h)}^2} \\ &= \frac{-2x}{x^2 x^2}\\ &= -\frac{2}{x^3} \end{align*} \]
Exercise 23.
\[ \begin{align*} \dd{f}{x} &= \lim_{h \to 0} \frac{\sqrt{x+h} - \surd x}{h} \\ &= \lim_{h \to 0} \frac{(\sqrt{x+h} - \surd x)(\sqrt{x+h}+\surd x)}{h(\sqrt{x+h}+\surd x)} \\ &= \lim_{h \to 0} \frac{x+h - x}{h(\sqrt{x+h} + \surd x)} \\ &= \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \surd x)} \\ &= \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \surd x} \\ &= \frac{1}{2\surd x} \end{align*} \]
Applications
Exercise 24.
a. using function composition we get
\[ (v \circ T)(t) = k e^{a b e^{-c t}} \]
b. using the chain rule we get
\[ (v \circ T)'(t) = (v' \circ T)(t) \cdot T'(t) = a k e^{a b e^{-ct}} \cdot \left(-b c e^{-ct} \right) \]
Exercise 25.
Using linear approximation of a function, and knowing that if \(f(x) = \ln x\) then \(f'(x) = \dfrac{1}{x}\), we obtain \(\Delta f(x) \approx f'(x) \Delta x = \dfrac{1}{x} \alpha x = \alpha\). Hence \(\Delta f(x)\) is constant and independent of the value \(f(x)\). Constant error is an assumption in many statistical tests, for example in the t-test or in ANOVA. Therefore, when the original data have proportional error, the logarithmically transformed data are more suitable for statistical testing.
Exercise 26.
Finding an extremum of \(g(x)\) means finding solutions to the equation \(g'(x)=0\). Hence, the Newton-Raphson method can be applied with the following iteration equation:
\[ x_{n+1} = x_n - \frac{g'(x_n)}{g''(x_n)} \]
Exercise 27.
Exercise 28. Sensitivity of logistic regression
a. We can use the fact that \(\frac{1}{1+e^{-x}} = \frac{e^{x}}{1 + e^{x}} = e^{x} (1+e^{x})^{-1}\). \[ \dd{\phi(x)}{t} = e^{x} (1+e^{x})^{-1} - e^{x} e^{x} (1 + e^{x})^{-2} = e^{x}(1 + e^{x})^{-1} \left(1 - e^{x}(1 + e^{x})^{-1} \right) = \phi(x)(1 - \phi(x)) \]
c. The derivative indicates how quickly the output probability changes with respect to changes in \(x\). This governs how sensitive the logistic regression model is to variations in the input score \(x\). A larger derivative magnitude implies greater sensitivity to changes in \(x\).
d. The maximum of \(\tdd{\phi(x)}{x} = \phi(x)(1 - \phi(x))\) is obtained when \(\phi(x) = \frac{1}{2}\), i.e. halfway between the intersections 0 and 1 with the \(x\)-axis. This happens when \(x=0\) and \(\tdd{\phi(x)}{x} = \frac{1}{4}\).
Exercise 29. Curve fitting
a. The least squares error \(E(a)\) is: $E(a)= _i \((Y_i−(a X_i + 10))^2\)
Substitute the given data points: \[ E(a) = (12 - (1 a + 10))^2 + (15 - (2 a + 10))^2 + (16 - (3 a + 10))^2 + (19 - (4 a + 10))^2 \]
Simplify: \[ \begin{align*} E(a) &= (2 − a)^2 + (5 − 2a)^2 + (6−3a)^2 + (9 − 4a)^2 E(a) &= 146 − 132 a + 30a^2 \end{align*} \]
b. \[ \dd{E(a)}{a} = 60 a − 132 \]
c. \[ \begin{align*} 60a − 132 &= 0 \\ 60a &= 132 \\ a &= \frac{132}{60} \\ a &= 2.2 \end{align*} \]
Once you have \(a\), calculate \(E(a)\) using the optimal \(a\) value:
\[ \begin{align*} E(2.2) &= 30(2.2)^2 − 132(2.2) + 146 \\ E(2.2) &= 145.2 − 290.4 + 146 \\ E(2.2) &= 0.8 \end{align*} \]
d. The slope \(a=2.2\) suggests that for every additional year in age, the gene expression level increases by approximately \(2.2\) TPM.
e. The RMSE value is \(\sqrt{0.8/4} = 0.447\). It represents the average deviation of the observed gene expression levels from the predicted gene expression levels.
5.4 Differential equations
Exercise 1.
a. In the equilibrium \(V'(a) = 0\), thus \(\alpha {V(a)}^\frac{2}{3}-\beta V(a) = 0\). If we solve this for \(V(a)\), we get \({(\frac{\alpha}{\beta})}^3\). Therefore \(V^* = {(\frac{\alpha}{\beta})}^3\)
b. The growth is maximal where \(V'(a)\) is maximal. Remember that you can find the maxima and minima by setting the derivative to 0. Because we want to know the value of \(V(a)\) for which \(V'(a)\) is maximal (and not the value of a!), we take the derivative of \(V'(a)\) to the function \(V'(a)\). \(\frac{V'(a)}{V(a)} = \frac{2}{3} \alpha {V(a)}^{-\frac{1}{3}} - \beta\). If we set this equal to 0, we get \(V(a) = \frac{8}{27} {(\frac{\alpha}{\beta})}^3\). This is a maximum (draw the graph or see that the second derivative (to the function \(V(a)\)!) is negative). Therefore \(V_{max}=\frac{8}{27} {(\frac{\alpha}{\beta})}^3\).
c.
Yes, if the volume (\(V\)) is lower than \(V*\), the derivative is positive and thus the volume will increase. If the volume (\(V\)) is higher than \(V*\), the derivative is negative and the volume will decrease.
Exercise 2.
a. Use rule 2: If the differential equation is of the form \(y'(t) = c_1 y(t) + c_2\), the solutions are of the form: \(y(t) = \left( \frac{c_2}{c_1} + y_0 \right) e^{c_1 t} - \frac{c_2}{c_1}\). Here \(c_1 = -b\) and \(c_2 = a\) and the solution is
\[ m(t) = \left(-\frac{a}{b} + m_0 \right) e^{-bt} + \frac{a}{b} \]
b. \(b = 0.1\) and \(m_0 = 2.5\). When \(t \to \infty\) then \(e^{-0.1t} \to 0\) and \(m(t) \to \frac{a}{b}\). Therefore, \(a\) should be high enough so that for large \(t\), \(m(t)\) is still at least 2. Therefore \(\frac{a}{b}\) has to be at least 2. Since \(b = 0.1\) we need \(a \geq 0.2\). You should add at least 0.2 \(\text{mg}\,\text{hr}^{-1}\).
Exercise 3.
a. \(X\) is produced by reaction 1 with rate \(k_1 [S]\) (rate constant times substrate) and by reaction 2 with \(k_{-2} [P]\) (this is the back reaction from the product, so here \(P\) is the substrate). \(X\) is consumed by reaction 1 going back to \(S\) with rate \(k_{-1} [X]\) and with reaction 2 going to \(P\) with rate \(k_{2} [X]\). This leads to the differential equation:
\[ \dd{[X]}{t} = k_1 [S] + k_{-2} [P] - (k_{-1} + k_2) [X] = 100 k_1 - (k_{-1} + k_2) [X] \]
b. \[ [X](t) = (\frac{100 k_1}{-(k_{-1} + k_2)} + X_0) e^{-(k_{-1} + k_2) t} - \frac{100 k_1}{-(k_{-1} + k_2)} = \frac{100 k_1}{-(k_{-1} + k_2)} e^{-(k_{-1} + k_2) t} + \frac{100 k_1}{k_{-1} + k_2} \]
c. \(t=\frac{\ln 2}{k_{-1}+k_2}\) (when \(t\rightarrow \infty\) (to reach steady state) \([X] \rightarrow \frac{100 k_1}{k_{-1} + k_2}\). To reach half that value \(e^{-(k_{-1} + k_2) t}\) has to equal \(\frac{1}{2}\). For that \(t=\frac{\ln(\frac{1}{2})}{-(k_{-1}+k_2)}\) which is equal to \(t=\frac{\ln 2}{k_{-1}+k_2}\).
Exercise 4.
a. The part \(a \left( 1 - \frac{x^n}{b^n + x^n} \right)\) gives the rate of synthesis and the part \(-kx\) gives the rate of decrease.
b. The rate of synthesis maximally equals \(a\) (when \(x=0\)) and minimally \(0\), when \(x \to \infty\).
d. For example by cooperative binding to the DNA by multimers of the transcription factor.
e. Negative autoregulation, negative feedback
Exercise 5.
a. The function intersects the \(X\)-axis at \(X=0\) and \(X=100\), and is an upside-down parabola with a maximum at \(X=0\). Therefore: \(\dd{X}{t}\), not \(X\), has a maximum at \(X=50\).
b. The stable steady state is at \(X=100\). When starting at \(X=1\), the population grows until 100 individuals, and when starting at 1000 it decreases until stabiilizing at 100.
c. The first equation can not be a solution because it has no asymptote. It grows to infinity. The second equation has an asymptote at 1 when \(t\) tends to large values. Therefore, it can also not be a solution. The third equation tends to 100 as \(t \to \infty\). It could be a solution. We will check that. Apply the quotient rule to find the left-hand side: \[ X'(t) = \frac{100 e^{-t}}{{(1 + e^{-t})}^2} \] Substituting the right-hand side of the differential equation yields \[ \begin{align*} X(t) \left( 1 - \frac{X(t)}{100} \right) & = \frac{100}{1 + e^{-t}} \left(1 - \frac{1}{1 + e^{-t}} \right) \\ & = \frac{100}{1 + e^{-t}} \left( \frac{1 + e^{-t}}{1 + e^{-t}} - \frac{1}{1 + e^{-t}} \right) \\ & = \frac{100}{1 + e^{-t}} \left( \frac{1 + e^{-t} - 1}{1 + e^{-t}} \right) \\ & = \frac{100 e^{-t}}{{(1 + e^{-t})}^2} \end{align*} \] Left- and right-hand sides are equal. Therefore, the third function is a solution to the differntial equation.
Exercise 6.
a. Intersections: \[ \begin{align*} \frac{5 x^2}{4 + x^2} - x & = \frac{5 x^2}{4 + x^2} - \frac{4x + x^3}{4 + x^2} \\ & = \frac{-x^3 + 5 x^2 - 4 x}{4 + x^2} = 0 \end{align*} \] Then \(-x(x^2 -5x +4) = -x(x-1)(x-4) = 0\), which has solutions: \(x=0\), \(x=1\) or \(x=4\).
b. \(x=0\), \(x=4\) and \(x=1\), respectively. The last starting point is an unstable steady state. Small deviations from it will let \(x\) go to either \(0\) or \(4\), the stable steady states.
c. That can not be a solution because it tends to \(1\) when \(t \to \infty\). We saw above that \(x=1\) is not a stable steady state, so a solution can not converge to that value unless it starts and remains there.
Metabolic networks
Exercise 7.
a. \[ \begin{align*} \frac{dB}{dt} &= v_1 - v_3 \\ \frac{dC}{dt} &= v_2 - v_4 \\ \frac{dD}{dt} &= v_3 + v_4 - v_5 \end{align*} \]
b. \[ v_1 = v_3, \quad v_2 = v_4. \]
c. \[ \begin{align*} \dd{B}{t} &= 100k_1 + k_{-3}[D] - (k_{-1} + k_3)[B] \\ \dd{C}{t} &= 100k_2 + k_{-4}[D] - (k_{-2} + k_4)[C] \\ \dd{D}{t} &= k_3[B] + k_4[C] - (k_{-3} + k_{-4} + k_5)[D] \end{align*} \]
Exercise 8.
a. \[ \dd{B}{t} = v_1 - v_2. \]
b.
c.
The steady state is when \(\frac{dB}{dt} = 0\), that is the intersection of \(v_1 - v_2\) with the \(x\)-axis. The intersection exists when \(v_2 > v_1\) for large \([B]\), \[ v_{\max 2} > \frac{v_{\max 1}[A]}{[A] + K_1} \quad \text{or specifically when} \quad v_{\max 2} > \frac{v_{\max 1}\cdot 2}{2 + K_1}. \]
d.
The SIR model
Exercise 9.
a. The basal reproduction number \(R_0 =\frac{\beta}{\gamma}\) equals \(R_\text{eff}\) when \(i\approx1\), so \(R_0 \approx 2.1--2.2\) at the start of the epidemic when no social restrictions were imposed yet.
b. When no social restrictions are imposed \(R_\text{eff} \approx 2.2 \cdot s\). Therefore, \(s(0) < 0.45\) to have \(R_\text{eff} < 1\) and \(r(0)>0.55\). So, 55% of the population should be resistant either due to vaccination or by having had the disease.
Exercise 10.
a. \(P_{i}(t) = e^{-\gamma t}\)
b. \(\defint{0}{\infty}{e^{-\gamma t}}{t} = {\left[\frac{-1}{\gamma} e^{-\gamma t} \right]}_{0}^{\infty} = \frac{1}{\gamma}\)
c. Clearly, that must be \(\defint{0}{\infty}{s \beta e^{-\gamma t}}{t}\) or \(\frac{\beta}{\gamma} = R_{0}\), when \(s \approx 1\) or \(\frac{s \beta}{\gamma} = R_\text{eff}\) otherwise.
Exercise 11.
a. \(\gamma \approx 0.5\) per week. Since \(R_0 \approx 2.2\) we obtain \(\beta \approx 1.1\) per week.
b. The peak fraction is approximately 20%, which is 3.4 million people. One percent of these is 340000 people.