Appendices

Appendices A to D of the book "Probability for the Enthusiastic Beginner" by David Morin (2016) Subtleties about probability Euler's number, e Approximations to (1 + a)^n The slope of e^x

65 flashcards · Mathematics · by lakesare

Part of Textbooks › Probability for the Enthusiastic Beginner (David Morin) · 7 courses

[Subtleties about probability] What are the two ways to determine the probability of an event?
Theoretical calculation Reason about the process, e.g. by noting which outcomes are equally likely. Example: Heads and Tails of a fair coin are equally likely, so P(Heads)=1/2P(\text{Heads}) = 1/2P(Heads)=1/2 . Experimental estimate Perform a large number of trials and take the fraction of trials in which the event occurred. Example: a bus is early on 20 of 50 days, so P(early)≈40%P(\text{early}) \approx 40\%P(early)≈40% .
[Subtleties about probability] Why can the probability that a bus arrives early, or that a basketball player makes a free throw, only be estimated experimentally?
Too many factors influence the outcome for a theoretical calculation to be trusted. Bus: traffic, weather, engine issues, the driver's sleep. Free throw: the player's thoughts, muscles, air currents.The best estimate is the fraction of past trials with the outcome. Such estimates should be taken with a grain of salt: conditions can change (a construction project can make the bus late every day for three months).
[Subtleties about probability] Is a coin toss truly random?
No. Once the coin leaves the hand, the side that lands facing up is completely determined by the force, torque, air, ground and coin. These quantities have definite values whether or not we know them. The "randomness" of the toss is only a result of our ignorance of them.
[Subtleties about probability] Why does a machine that flips a coin with a random force and torque not make the toss truly random?
The force and torque would themselves have to be random, so the question of how to produce randomness has only moved back one step.
[Subtleties about probability] Why can a computer's random number generator not produce truly random numbers?
Its output is completely determined by its input, and the input is not random (otherwise no random number generator would be needed). The numbers only seem random.
[Subtleties about probability] Definition of probability: in a very large number NNN of identical trials, an event of probability ppp occurs in a fraction ppp of the trials, on average.Why does this definition seem to rule out true randomness, and what resolves this?
Truly identical trials should give identical results. Quantum mechanics resolves this: there, exactly identical processes can give different results. In the coin-toss scenarios the trials are never identical: they have slightly different inputs. In quantum mechanics some things happen without a cause.
[Subtleties about probability] Without quantum mechanics, what two ingredients make many physical processes essentially random?
Large numbers An enormous number of particles or inputs is involved. Example: the radio noise in the atmosphere, used by random.org to generate random numbers. Chaos Small uncertainties grow into huge ones. Example: the ping-pong balls bouncing in a lottery box.
[Subtleties about probability] I flip a coin, see that it shows Heads, and do not show it to you. What is the probability that the coin shows Heads?
1/21/21/2 for you, 111 for me. The question is incomplete: it must state who calculates the probability. The two processes differ: my trials include looking at the coin, so to me they are not identical. 1/21/21/2 is the probability with no other information; 111 is the conditional probability of Heads given that Heads was observed.
[Subtleties about probability] Definition of probability: in a very large number NNN of identical trials, an event of probability ppp occurs in a fraction ppp of the trials, on average.Why can the words "on average" not be dropped from this definition?
Nothing guarantees that the event occurs in exactly a fraction ppp of the trials. Requiring only a fraction in a small interval around ppp does not help: nothing guarantees ending up in that interval either.
[Subtleties about probability] Definition of probability: in a very large number NNN of identical trials, an event of probability ppp occurs in a fraction ppp of the trials, on average.Why does "on average" make this definition circular?
The average over N′N'N′ sets of NNN trials equals ppp only "on average" again, which needs N′′N''N′′ sets of N′N'N′ sets, and so on without end. Every statement we can make ends with "on average", so we can never be certain of getting within, say, 0.00010.00010.0001 of ppp . Morin: "a never-ending linear chain" describes the definition better than "circular".
[Subtleties about probability] Does regarding N′′N''N′′ sets of N′N'N′ sets of NNN trials as one single set of N′′N′NN'' N' NN′′N′N trials remove the endless "on average" regress in the definition of probability?
No. Any statement about that single set still ends with "on average". So we again need N′′′N'''N′′′ sets of that many trials, however the trials are grouped.
[Subtleties about probability] About how likely is it that a trillion flips of a fair coin give a Heads percentage outside 49.999%49.999\%49.999% to 50.001%50.001\%50.001% ?
About 5⋅10−895 \cdot 10^{-89}5⋅10−89 . The same holds for a million flips and the range 49%49\%49% to 51%51\%51% . Both ranges extend 20 standard deviations on each side of 50%50\%50% : e.g. for 10610^{6}106 flips, σ=0.5/106=0.05%\sigma = 0.5/\sqrt{10^{6}} = 0.05\%σ=0.5/106=0.05% of the flips, and 1%=20 σ1\% = 20\,\sigma1%=20σ .
[Subtleties about probability] Can any number of trials make you absolutely sure that a fair coin gives Heads half the time?
No. However many trials you do, you cannot be absolutely sure that you have not simply had bad (or good) luck. No large number marks a point where things turn from probable to definite: in that sense 10100010^{1000}101000 is no better than 101010 ; the difference is only one of degree.
[Subtleties about probability] Given these philosophical problems with the foundations of probability, what is Morin's practical conclusion?
Keep using probability: in practice it works. The book takes the practical approach of intuitively knowing that P(Heads)=1/2P(\text{Heads}) = 1/2P(Heads)=1/2 and P(rolling a 5)=1/6P(\text{rolling a 5}) = 1/6P(rolling a 5)=1/6 .
[Euler's number, e] What is the definition of Euler's number eee ?
e≡lim⁡n→∞(1+1n)n≈2.71828\boldsymbol{e \equiv \displaystyle\lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^{n} \approx 2.71828}e≡n→∞lim(1+n1)n≈2.71828 Example: (1+1n)n\left(1 + \tfrac{1}{n}\right)^{n}(1+n1)n for n=1,2,5,10,100,1000n = 1, 2, 5, 10, 100, 1000n=1,2,5,10,100,1000 : 2, 2.25, 2.49, 2.59, 2.705, 2.7172, \ 2.25, \ 2.49, \ 2.59, \ 2.705, \ 2.7172, 2.25, 2.49, 2.59, 2.705, 2.717 (1+1/n)n(1 + 1/n)^n(1+1/n)n plotted against log⁡10n\log_{10} nlog10n ; the dashed line is at eee .
[Euler's number, e] What are the first 10 digits of eee ?
2.718281828\boldsymbol{2.718281828}2.718281828 "1828" appears twice, which makes the digits easy to remember. The digits continue forever with no overall pattern.
[Euler's number, e] As n→∞n \to \inftyn→∞ , why do neither " (1+1/n)n→1(1 + 1/n)^n \to 1(1+1/n)n→1 " nor " (1+1/n)n→∞(1 + 1/n)^n \to \infty(1+1/n)n→∞ " hold?
The two effects cancel: the base 1+1/n1 + 1/n1+1/n tends to 1, but the exponent nnn tends to ∞\infty∞ ; the limit is e≈2.71828e \approx 2.71828e≈2.71828 , between 1 and ∞\infty∞ . 111 raised to any power is 1, which suggests the limit 1; a power that grows without bound suggests the limit ∞\infty∞ .
[Euler's number, e] Why is (1+1/n)n(1 + 1/n)^n(1+1/n)n plotted against log⁡10n\log_{10} nlog10n instead of nnn ?
On a linear nnn axis up to n=104n = 10^4n=104 , the curve looks like a vertical jump from 2.0 to 2.718 followed by a horizontal line. The region of small nnn is squeezed to almost nothing (up to n=100n = 100n=100 is only 0.01%0.01\%0.01% of a plot up to 10610^6106 ). A log axis keeps large numbers from dominating the plot.
[Euler's number, e] Does the equation e=lim⁡n→∞(1+1/n)ne = \displaystyle\lim_{n\to\infty} (1 + 1/n)^ne=n→∞lim(1+1/n)n state a result?
No, it is a definition: it gives the name eee to the value that (1+1/n)n(1 + 1/n)^n(1+1/n)n approaches as nnn becomes large. If you do not like limits, set nnn to a very large number like 101010^{10}1010 : (1+10−10)1010(1 + 10^{-10})^{10^{10}}(1+10−10)1010 is essentially eee .
[Euler's number, e] Write exe^xex as a limit.
ex=lim⁡n→∞(1+xn)n\boldsymbol{e^x = \displaystyle\lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^{n}}ex=n→∞lim(1+nx)n So ex≈(1+x/n)ne^x \approx (1 + x/n)^nex≈(1+x/n)n for large nnn . The case x=1x = 1x=1 is the definition of eee .
[Euler's number, e] e=lim⁡m→∞(1+1m)me = \displaystyle\lim_{m\to\infty} \left(1 + \frac{1}{m}\right)^{m}e=m→∞lim(1+m1)m How does ex=lim⁡n→∞(1+xn)ne^x = \displaystyle\lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^{n}ex=n→∞lim(1+nx)n follow from this?
ex=lim⁡m→∞(1+1m)mx=lim⁡m→∞(1+xmx)mxe^x = \displaystyle\lim_{m\to\infty} \left(1 + \frac{1}{m}\right)^{mx} = \lim_{m\to\infty} \left(1 + \frac{x}{mx}\right)^{mx}ex=m→∞lim(1+m1)mx=m→∞lim(1+mxx)mx , then set n=mxn = mxn=mx . First step: raise to the power xxx and multiply exponents. Second step: multiply 1/m1/m1/m by 111 in the form x/xx/xx/x .
[Euler's number, e] Under what condition on xxx and nnn is ex≈(1+xn)ne^x \approx \left(1 + \frac{x}{n}\right)^{n}ex≈(1+nx)n a good approximation?
x≪n\boldsymbol{x \ll \sqrt{n}}x≪n Example: x=2x = 2x=2 , n=106n = 10^6n=106 : (1+2/106)106=7.389041(1 + 2/10^6)^{10^6} = 7.389041(1+2/106)106=7.389041 , while e2≈7.389056e^2 \approx 7.389056e2≈7.389056 .
[Euler's number, e] A bank account pays 5%5\%5% per year, compounded daily: each day the money is multiplied by 1+0.05/3651 + 0.05/3651+0.05/365 . By what factor does the money grow in one year?
(1+0.05365)365≈e0.05≈1.051\boldsymbol{\left(1 + \frac{0.05}{365}\right)^{365} \approx e^{0.05} \approx 1.051}(1+3650.05)365≈e0.05≈1.051 This has the form (1+x/n)n(1 + x/n)^n(1+x/n)n with x=0.05x = 0.05x=0.05 and n=365n = 365n=365 , and x≪nx \ll \sqrt{n}x≪n , so it is close to exe^xex .
[Euler's number, e] What is the yield of an interest rate?
The effective yearly interest rate: the one-time end-of-year rate that gives the same amount of money. It exceeds the annual rate because interest is also earned on the interest already received.Example: 5%5\%5% compounded daily gives the factor e0.05≈1.051e^{0.05} \approx 1.051e0.05≈1.051 , so a yield of 5.1%5.1\%5.1% .Example: 10%10\%10% gives e0.10≈1.105e^{0.10} \approx 1.105e0.10≈1.105 , a yield of 10.5%10.5\%10.5% ; 20%20\%20% gives e0.20≈1.22e^{0.20} \approx 1.22e0.20≈1.22 , a yield of 22…
[Euler's number, e] At 5%5\%5% per year compounded daily, how many years does it take to double your money?
About 14 years. After nnn years the factor is (e0.05)n=e0.05n(e^{0.05})^n = e^{0.05n}(e0.05)n=e0.05n ; e0.7≈2e^{0.7} \approx 2e0.7≈2 (i.e. ln⁡2≈0.7\ln 2 \approx 0.7ln2≈0.7 ), so 0.05n=0.70.05n = 0.70.05n=0.7 and n=14n = 14n=14 .
[Euler's number, e] Why is "the money grows by about 0.05N0.05N0.05N dollars per year, so doubling NNN dollars takes 1/0.05=201/0.05 = 201/0.05=20 years" wrong?
It ignores compound interest: each year there is more money, so the interest is earned on a larger and larger amount. The correct answer is about 14 years. Over 20 years the interest earned is large (about NNN ), so interest on interest cannot be neglected.
[Euler's number, e] Rule of 70: at an interest rate of r%r\%r% per year, about how many years does it take to double your money?
70/r\boldsymbol{70/r}70/r years Because ln⁡2≈0.70\ln 2 \approx 0.70ln2≈0.70 . Equivalently, 70%70\%70% of the naive answer 100/r100/r100/r that ignores compound interest.Example: at 10%10\%10% , the money doubles in about 7 years.
[Euler's number, e] At 5%5\%5% per year applied only once at the end of each year (no daily compounding), how long does it take to double your money?
Still essentially 14 years: (1.05)14≈1.98≈2(1.05)^{14} \approx 1.98 \approx 2(1.05)14≈1.98≈2 . The extra 0.1%0.1\%0.1% per year from daily compounding makes little difference.
[Euler's number, e] What is the infinite series (Taylor series) for exe^xex ?
ex=1+x+x22!+x33!+x44!+⋯\boldsymbol{e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \cdots}ex=1+x+2!x2+3!x3+4!x4+⋯ Every term has the form xn/n!x^n/n!xn/n! , with nnn from 000 to ∞\infty∞ .Example: at x=1x = 1x=1 : e=1+1+12!+13!+⋯e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdotse=1+1+2!1+3!1+⋯ ; the terms up to 1/10!1/10!1/10! give e≈2.71828180e \approx 2.71828180e≈2.71828180 , correct to 7 decimal places.
[Euler's number, e] What is the simplest approximation of exe^xex for small xxx that depends on xxx , and why does it hold?
ex≈1+x\boldsymbol{e^x \approx 1 + x}ex≈1+x In ex=1+x+x2/2!+⋯e^x = 1 + x + x^2/2! + \cdotsex=1+x+x2/2!+⋯ , the terms x2/2!x^2/2!x2/2! and higher are small compared with xxx .Example: e0.01=1.01005≈1.01e^{0.01} = 1.01005 \approx 1.01e0.01=1.01005≈1.01 .
[Euler's number, e] Does bx≈1+xb^x \approx 1 + xbx≈1+x for small xxx hold for bases bbb other than eee ?
No, eee is the only such base. Example: 20.01≈1.00702^{0.01} \approx 1.007020.01≈1.0070 and 100.01≈1.023310^{0.01} \approx 1.0233100.01≈1.0233 , neither close to 1.011.011.01 ; but e0.01≈1.0101e^{0.01} \approx 1.0101e0.01≈1.0101 .
[Euler's number, e] For small xxx , which approximations of exe^xex are there, and how do you choose between them?
ex≈1, ex≈1+x, ex≈1+x+x2/2,…\boldsymbol{e^x \approx 1, \ \ e^x \approx 1 + x, \ \ e^x \approx 1 + x + x^2/2, \dots}ex≈1, ex≈1+x, ex≈1+x+x2/2,… (keeping more terms of the series); use the coarsest one that is accurate enough for your purpose. ex≈1e^x \approx 1ex≈1 drops xxx and all higher powers, which are small compared with 1. In many cases ex≈1+xe^x \approx 1 + xex≈1+x is enough.
[Euler's number, e] How does the series ex=∑kxk/k!e^x = \sum_{k} x^k/k!ex=∑kxk/k! follow from ex=lim⁡n→∞(1+x/n)ne^x = \displaystyle\lim_{n\to\infty} (1 + x/n)^nex=n→∞lim(1+x/n)n ?
Expand (1+x/n)n(1 + x/n)^n(1+x/n)n with the binomial theorem: the xkx^kxk term is (nk)xknk=xkk!⋅n(n−1)⋯(n−k+1)nk\displaystyle\binom{n}{k} \frac{x^k}{n^k} = \frac{x^k}{k!} \cdot \frac{n(n-1)\cdots(n-k+1)}{n^k}(kn)nkxk=k!xk⋅nkn(n−1)⋯(n−k+1) , and the last factor tends to 1. ex=lim⁡n→∞[1+x(nn)+x22!(n(n−1)n2)+x33!(n(n−1)(n−2)n3)+⋯ ]e^x = \displaystyle\lim_{n\to\infty} \left[1 + x\left(\frac{n}{n}\right) + \frac{x^2}{2!}\left(\frac{n(n-1)}{n^2}\right) + \frac{x^3}{3!}\left(\frac{n(n-1)(n-2)}{n^3…
[Euler's number, e] Why does n(n−1)(n−2)n3→1\dfrac{n(n-1)(n-2)}{n^3} \to 1n3n(n−1)(n−2)→1 as n→∞n \to \inftyn→∞ ?
The ratios (n−1)/n(n-1)/n(n−1)/n and (n−2)/n(n-2)/n(n−2)/n both tend to 1: for large nnn , n−1n - 1n−1 and n−2n - 2n−2 are essentially equal to nnn in a multiplicative sense. For large but finite nnn , the factors far out in the series differ from 1, but there the factorials in the denominators make the terms negligible.
[Euler's number, e] How does the decimal expansion e0.0001=1.000100005000166…e^{0.0001} = 1.000100005000166\ldotse0.0001=1.000100005000166… show the first coefficients of the series for exe^xex ?
It splits as 1+0.0001+0.000000005+0.000000000000166…=1+x+x22!+x33!+⋯1 + 0.0001 + 0.000000005 + 0.000000000000166\ldots = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots1+0.0001+0.000000005+0.000000000000166…=1+x+2!x2+3!x3+⋯ with x=0.0001x = 0.0001x=0.0001 . So if ex=a+bx+cx2+dx3+⋯e^x = a + bx + cx^2 + dx^3 + \cdotsex=a+bx+cx2+dx3+⋯ , then a=b=1a = b = 1a=b=1 , c=1/2c = 1/2c=1/2 , d=1/6d = 1/6d=1/6 .
[Euler's number, e] What is the slope of the curve exe^xex at a point xxx ?
ex\boldsymbol{e^x}ex , the value of the function at that point. The slope at a point is the slope of the tangent line there. eee is the only base with this property: it fails for 2x2^x2x and 10x10^x10x . Example: slope 1/e1/e1/e at x=−1x = -1x=−1 , slope e0=1e^0 = 1e0=1 at x=0x = 0x=0 , slope eee at x=1x = 1x=1 , slope e2≈7.39e^2 \approx 7.39e2≈7.39 at x=2x = 2x=2 .
[Euler's number, e] Why does AexAe^xAex ( AAA a constant) also have slope equal to its value at every point?
Both the value and the slope of AexAe^xAex are AAA times those of exe^xex , and for exe^xex they are equal.
[Approximations to (1 + a)^n] For small aaa , what is the basic approximation of (1+a)n(1 + a)^n(1+a)n ?
(1+a)n≈ena\boldsymbol{(1 + a)^n \approx e^{na}}(1+a)n≈ena It is ex≈(1+x/n)ne^x \approx (1 + x/n)^nex≈(1+x/n)n with a=x/na = x/na=x/n . It is used in deriving the exponential and Poisson distributions.
[Approximations to (1 + a)^n] What is the more accurate approximation of (1+a)n(1 + a)^n(1+a)n needed for the Gaussian approximations to the binomial and Poisson distributions?
(1+a)n≈enae−na2/2\boldsymbol{(1 + a)^n \approx e^{na} e^{-na^2/2}}(1+a)n≈enae−na2/2 When aaa is small enough, e−na2/2≈e−0=1e^{-na^2/2} \approx e^{-0} = 1e−na2/2≈e−0=1 , and this reduces to (1+a)n≈ena(1 + a)^n \approx e^{na}(1+a)n≈ena .
[Approximations to (1 + a)^n] With n=100n = 100n=100 and a=1/10a = 1/10a=1/10 , compare (1+a)n(1 + a)^n(1+a)n with enae^{na}ena and with enae−na2/2e^{na} e^{-na^2/2}enae−na2/2 .
(1.1)100≈13,781\boldsymbol{(1.1)^{100} \approx 13{,}781}(1.1)100≈13,781 ; e10≈22,026\boldsymbol{e^{10} \approx 22{,}026}e10≈22,026 (a very bad approximation); e10e−1/2≈13,360\boldsymbol{e^{10} e^{-1/2} \approx 13{,}360}e10e−1/2≈13,360 (about 3%3\%3% off). e−1/2≈0.60653e^{-1/2} \approx 0.60653e−1/2≈0.60653 and 22,026⋅0.60653≈13,36022{,}026 \cdot 0.60653 \approx 13{,}36022,026⋅0.60653≈13,360 .
[Approximations to (1 + a)^n] In what sense does Morin use " ≈\approx≈ "?
Multiplicative: the ratio of the two numbers is close to 1, not their difference close to 0. Example: 13,36013{,}36013,360 and 13,78113{,}78113,781 differ by 421, yet their ratio is 0.970.970.97 , so 13,360≈13,78113{,}360 \approx 13{,}78113,360≈13,781 .
[Approximations to (1 + a)^n] What exact formula for (1+a)n(1 + a)^n(1+a)n ( ∣a∣<1|a| < 1∣a∣<1 ) do the approximations enae^{na}ena and enae−na2/2e^{na}e^{-na^2/2}enae−na2/2 come from?
(1+a)n=ena e−na2/2 ena3/3 e−na4/4 ena5/5⋯\boldsymbol{(1 + a)^n = e^{na} \, e^{-na^2/2} \, e^{na^3/3} \, e^{-na^4/4} \, e^{na^5/5} \cdots}(1+a)n=enae−na2/2ena3/3e−na4/4ena5/5⋯ Exact with infinitely many factors. The approximations keep the first one or two factors.
[Approximations to (1 + a)^n] How does the derivation of (1+a)n=enae−na2/2ena3/3⋯(1 + a)^n = e^{na} e^{-na^2/2} e^{na^3/3} \cdots(1+a)n=enae−na2/2ena3/3⋯ (for ∣a∣<1|a| < 1∣a∣<1 ) go?
Geometric series 1−a+a2−a3+a4−⋯=11+a1 - a + a^2 - a^3 + a^4 - \cdots = \dfrac{1}{1 + a}1−a+a2−a3+a4−⋯=1+a1 Integrate both sides with respect to aaa a−a22+a33−a44+⋯=ln⁡(1+a)a - \dfrac{a^2}{2} + \dfrac{a^3}{3} - \dfrac{a^4}{4} + \cdots = \ln(1 + a)a−2a2+3a3−4a4+⋯=ln(1+a) Exponentiate both sides Use eln⁡(1+a)=1+ae^{\ln(1+a)} = 1 + aeln(1+a)=1+a and that the exponential of a sum is the product of the exponentials. 1+a=eae−a2/2ea3/3e−a4/4⋯1 + a = e^{a} e^{-a^2/2} e^{a^3/3} e^{-a^4/4} \cdots1+a=e…
[Approximations to (1 + a)^n] How can you check 1−a+a2−a3+⋯=11+a1 - a + a^2 - a^3 + \cdots = \dfrac{1}{1 + a}1−a+a2−a3+⋯=1+a1 (for ∣a∣<1|a| < 1∣a∣<1 )?
Multiply both sides by 1+a1 + a1+a : on the left the infinitely many cross terms cancel in pairs, and only the 1 is left. For ∣a∣≥1|a| \ge 1∣a∣≥1 the formula is false: a=2a = 2a=2 gives 1−2+4−8+⋯1 - 2 + 4 - 8 + \cdots1−2+4−8+⋯ on the left but 1/31/31/3 on the right.
[Approximations to (1 + a)^n] a−a22+a33−⋯=ln⁡(1+a)a - \dfrac{a^2}{2} + \dfrac{a^3}{3} - \cdots = \ln(1 + a)a−2a2+3a3−⋯=ln(1+a) is obtained by integrating 1−a+a2−⋯=11+a1 - a + a^2 - \cdots = \dfrac{1}{1 + a}1−a+a2−⋯=1+a1 .Why is the constant of integration zero?
At a=0a = 0a=0 both sides are 0 ( ln⁡1=0\ln 1 = 0ln1=0 ). This is the Taylor series for ln⁡(1+a)\ln(1 + a)ln(1+a) , valid for ∣a∣<1|a| < 1∣a∣<1 .
[Approximations to (1 + a)^n] (1+a)n=ena e−na2/2 ena3/3 e−na4/4⋯(1 + a)^n = e^{na} \, e^{-na^2/2} \, e^{na^3/3} \, e^{-na^4/4} \cdots(1+a)n=enae−na2/2ena3/3e−na4/4⋯ for ∣a∣<1|a| < 1∣a∣<1 For a≪1a \ll 1a≪1 , which approximation of (1+a)n(1 + a)^n(1+a)n holds under which condition on nnn and aaa ?
na≪1na \ll 1na≪1 : (1+a)n≈1\ (1 + a)^n \approx 1 (1+a)n≈1 Example: n=1n = 1n=1 , a=1/100a = 1/100a=1/100 : (1.01)1=1.01(1.01)^1 = 1.01(1.01)1=1.01 , only 1%1\%1% above 1. na2≪1na^2 \ll 1na2≪1 : (1+a)n≈ena\ (1 + a)^n \approx e^{na} (1+a)n≈ena Example: n=100n = 100n=100 , a=1/100a = 1/100a=1/100 : e1=2.718e^{1} = 2.718e1=2.718 , while (1.01)100≈2.705(1.01)^{100} \approx 2.705(1.01)100≈2.705 ( 0.5%0.5\%0.5% off). na3≪1na^3 \ll 1na3≪1 : (1+a)n≈enae−na2/2\ (1 + a)^n \approx e^{na} e^{-na^2/2} (1+a)n…
[Approximations to (1 + a)^n] (1+a)n=ena e−na2/2 ena3/3⋯(1 + a)^n = e^{na} \, e^{-na^2/2} \, e^{na^3/3} \cdots(1+a)n=enae−na2/2ena3/3⋯ for ∣a∣<1|a| < 1∣a∣<1 If a≪1a \ll 1a≪1 and na≪1na \ll 1na≪1 , why does (1+a)n≈1(1 + a)^n \approx 1(1+a)n≈1 ?
All exponents are much smaller than 1, so every factor is ≈e0=1\approx e^0 = 1≈e0=1 . nanana is small by assumption; na2/2na^2/2na2/2 is smaller than nanana by the factor a/2a/2a/2 ; each later exponent has further factors of aaa .
[Approximations to (1 + a)^n] (1+a)n≈ena(1 + a)^n \approx e^{na}(1+a)n≈ena holds when na2≪1na^2 \ll 1na2≪1 .What do you get for n=1n = 1n=1 ?
1+a≈ea\boldsymbol{1 + a \approx e^{a}}1+a≈ea , the approximation ex≈1+xe^x \approx 1 + xex≈1+x for small xxx .
[Approximations to (1 + a)^n] (1+a)n≈ena(1 + a)^n \approx e^{na}(1+a)n≈ena holds when na2≪1na^2 \ll 1na2≪1 .Where does the condition x≪nx \ll \sqrt{n}x≪n for ex≈(1+x/n)ne^x \approx (1 + x/n)^nex≈(1+x/n)n come from?
Set a=x/na = x/na=x/n : na2≪1na^2 \ll 1na2≪1 becomes n(x/n)2≪1n(x/n)^2 \ll 1n(x/n)2≪1 , i.e. x2≪nx^2 \ll nx2≪n , i.e. x≪n\boldsymbol{x \ll \sqrt{n}}x≪n .
[Approximations to (1 + a)^n] (1+a)n=ena e−na2/2 ena3/3 e−na4/4 ena5/5 e−na6/6 ena7/7⋯(1 + a)^n = e^{na} \, e^{-na^2/2} \, e^{na^3/3} \, e^{-na^4/4} \, e^{na^5/5} \, e^{-na^6/6} \, e^{na^7/7} \cdots(1+a)n=enae−na2/2ena3/3e−na4/4ena5/5e−na6/6ena7/7⋯ For n=1012n = 10^{12}n=1012 and a=1/100a = 1/100a=1/100 , which factors must be kept?
All factors up to e−na6/6e^{-na^6/6}e−na6/6 ; ena7/7e^{na^7/7}ena7/7 and later can be dropped. na6=1012⋅10−12=1na^6 = 10^{12} \cdot 10^{-12} = 1na6=1012⋅10−12=1 is not small, but na7=0.01≪1na^7 = 0.01 \ll 1na7=0.01≪1 . In general, keep factors until nak≪1na^k \ll 1nak≪1 .
[Approximations to (1 + a)^n] When (1+a)n=enae−na2/2ena3/3⋯(1 + a)^n = e^{na} e^{-na^2/2} e^{na^3/3} \cdots(1+a)n=enae−na2/2ena3/3⋯ is cut off after some factor, what is the approximate size of the multiplicative error?
The first dropped factor. All later factors are even closer to e0=1e^0 = 1e0=1 . If the first dropped exponent is positive, the true value is larger than the approximation; if negative, smaller.Example: n=10,000n = 10{,}000n=10,000 , a=1/100a = 1/100a=1/100 , keeping enae−na2/2e^{na}e^{-na^2/2}enae−na2/2 : the first dropped factor is ena3/3=e1/300≈1.0033e^{na^3/3} = e^{1/300} \approx 1.0033ena3/3=e1/300≈1.0033 , and indeed 1.6358/1.6304≈1.00331.6358/1.6304 \approx 1.00331.6358/1.6304≈1.0033 .
[Approximations to (1 + a)^n] (1+a)n=ena e−na2/2 ena3/3 e−na4/4⋯(1 + a)^n = e^{na} \, e^{-na^2/2} \, e^{na^3/3} \, e^{-na^4/4} \cdots(1+a)n=enae−na2/2ena3/3e−na4/4⋯ For n=10,000n = 10{,}000n=10,000 and a=1/100a = 1/100a=1/100 , how accurate is (1+a)n≈enae−na2/2ena3/3(1 + a)^n \approx e^{na} e^{-na^2/2} e^{na^3/3}(1+a)n≈enae−na2/2ena3/3 ?
Off by the factor e−na4/4=e−1/40,000≈0.999975e^{-na^4/4} = e^{-1/40{,}000} \approx 0.999975e−na4/4=e−1/40,000≈0.999975 , i.e. only 0.0025%0.0025\%0.0025% . The true value is the smaller one, since the first dropped exponent is negative.
[Approximations to (1 + a)^n] Why derive (1+a)n≈enae−na2/2(1 + a)^n \approx e^{na} e^{-na^2/2}(1+a)n≈enae−na2/2 with calculus rather than without?
Calculus gives approximations of any accuracy at once. The non-calculus derivation is laborious and becomes intractable when more factors are wanted.
[The slope of e^x] What is the strategy for finding the slope of a function f(x)f(x)f(x) at a point xxx ?
Take the average slope (rise over run) between xxx and a nearby point, then move the nearby point closer and closer to xxx . The average slopes approach the slope of the tangent line at xxx .
[The slope of e^x] What are the average slopes of exe^xex from x=2x = 2x=2 to x=2.1x = 2.1x=2.1 , to 2.012.012.01 , and to 2.0012.0012.001 , and what do they approach?
≈7.77, 7.43, 7.393\boldsymbol{\approx 7.77, \ 7.43, \ 7.393}≈7.77, 7.43, 7.393 , approaching 7.389=e27.389 = e^27.389=e2 . e2.1−e20.1≈7.77\dfrac{e^{2.1} - e^2}{0.1} \approx 7.770.1e2.1−e2≈7.77 . The average slopes are larger than the true slope because the curve exe^xex bends upward.
[The slope of e^x] What is the definition of the slope of f(x)f(x)f(x) at the value xxx ?
slope=lim⁡δ→0f(x+δ)−f(x)δ\boldsymbol{\text{slope} = \displaystyle\lim_{\delta \to 0} \frac{f(x + \delta) - f(x)}{\delta}}slope=δ→0limδf(x+δ)−f(x) The average slope between xxx and x+δx + \deltax+δ , as δ\deltaδ becomes smaller and smaller.Example: f(x)=x2f(x) = x^2f(x)=x2 : (x+δ)2−x2δ=2x+δ→2x\dfrac{(x+\delta)^2 - x^2}{\delta} = 2x + \delta \to 2xδ(x+δ)2−x2=2x+δ→2x .
[The slope of e^x] What is the first derivation that the slope of exe^xex equals exe^xex ?
ex+δ−exδ=ex eδ−1δ≈ex (1+δ)−1δ=ex\boldsymbol{\dfrac{e^{x+\delta} - e^x}{\delta} = e^x \, \dfrac{e^\delta - 1}{\delta} \approx e^x \, \dfrac{(1 + \delta) - 1}{\delta} = e^x}δex+δ−ex=exδeδ−1≈exδ(1+δ)−1=ex Factor out exe^xex , then use eδ≈1+δe^\delta \approx 1 + \deltaeδ≈1+δ for small δ\deltaδ . In the limit δ→0\delta \to 0δ→0 the " ≈\approx≈ " becomes " === ".
[The slope of e^x] Why does the derivation " ex+δ−exδ=exeδ−1δ≈ex\dfrac{e^{x+\delta} - e^x}{\delta} = e^x \dfrac{e^\delta - 1}{\delta} \approx e^xδex+δ−ex=exδeδ−1≈ex " not show that the slope of 2x2^x2x is 2x2^x2x ?
2δ≈1+δ2^\delta \approx 1 + \delta2δ≈1+δ is false for small δ\deltaδ ; eδ≈1+δe^\delta \approx 1 + \deltaeδ≈1+δ holds only for the base eee . Example: 20.01≈1.00702^{0.01} \approx 1.007020.01≈1.0070 , not 1.011.011.01 .
[The slope of e^x] In lim⁡δ→0f(x+δ)−f(x)δ\displaystyle\lim_{\delta \to 0} \frac{f(x + \delta) - f(x)}{\delta}δ→0limδf(x+δ)−f(x) there is a δ\deltaδ in the denominator. Why is this not a division by zero?
The δ\deltaδ in the denominator is cancelled against the numerator first; then the limit δ→0\delta \to 0δ→0 is taken, which is trivial when no δ\deltaδ is left in a denominator. Example: 2xδ+δ2δ=2x+δ→2x\dfrac{2x\delta + \delta^2}{\delta} = 2x + \delta \to 2xδ2xδ+δ2=2x+δ→2x .
[The slope of e^x] What is the slope of xnx^nxn ( nnn a nonnegative integer)?
nxn−1\boldsymbol{nx^{n-1}}nxn−1 It holds for any nnn , not only nonnegative integers. Check: n=0n = 0n=0 gives slope 0, n=1n = 1n=1 gives slope 1.Example: (x+δ)3−x3δ=3x2+3xδ+δ2→3x2\dfrac{(x+\delta)^3 - x^3}{\delta} = 3x^2 + 3x\delta + \delta^2 \to 3x^2δ(x+δ)3−x3=3x2+3xδ+δ2→3x2 .
[The slope of e^x] How does the proof that the slope of xnx^nxn is nxn−1nx^{n-1}nxn−1 go, for a nonnegative integer nnn ?
Binomial expansion: (x+δ)n=xn+(n1)xn−1δ+(n2)xn−2δ2+⋯(x + \delta)^n = x^n + \binom{n}{1} x^{n-1}\delta + \binom{n}{2} x^{n-2}\delta^2 + \cdots(x+δ)n=xn+(1n)xn−1δ+(2n)xn−2δ2+⋯ ; subtract xnx^nxn , divide by δ\deltaδ , and every term except (n1)xn−1=nxn−1\binom{n}{1}x^{n-1} = nx^{n-1}(1n)xn−1=nxn−1 still contains δ\deltaδ , so it tends to 0. Example: (x+δ)4=x4+(4x3)δ+6x2δ2+4xδ3+δ4(x+\delta)^4 = x^4 + (4x^3)\delta + 6x^2\delta^2 + 4x\delta^3 + \delta^4(x+δ)4=x4+(4x3)δ+6x2δ2+4xδ3+δ4 gives slope 4…
[The slope of e^x] Why is the slope of F(x)=f1(x)+f2(x)F(x) = f_1(x) + f_2(x)F(x)=f1(x)+f2(x) the sum of the slopes of f1f_1f1 and f2f_2f2 ?
In F(x+δ)−F(x)δ\dfrac{F(x+\delta) - F(x)}{\delta}δF(x+δ)−F(x) , group the f1f_1f1 terms and the f2f_2f2 terms: it equals f1(x+δ)−f1(x)δ+f2(x+δ)−f2(x)δ\dfrac{f_1(x+\delta) - f_1(x)}{\delta} + \dfrac{f_2(x+\delta) - f_2(x)}{\delta}δf1(x+δ)−f1(x)+δf2(x+δ)−f2(x) , and the limit of each part is a slope. The same works for any number of functions, even infinitely many (as needed for a series).
[The slope of e^x] What is the slope of AxnAx^nAxn , for a constant AAA ?
Anxn−1\boldsymbol{Anx^{n-1}}Anxn−1 The constant AAA factors out of A(x+δ)n−Axnδ\dfrac{A(x+\delta)^n - Ax^n}{\delta}δA(x+δ)n−Axn .Example: the slope of x3/3!x^3/3!x3/3! is 3x2/3!=x2/2!3x^2/3! = x^2/2!3x2/3!=x2/2! .
[The slope of e^x] How does the series ex=1+x+x22!+x33!+⋯e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdotsex=1+x+2!x2+3!x3+⋯ show that the slope of exe^xex is exe^xex ?
The slope of each term xn/n!x^n/n!xn/n! is nxn−1/n!=xn−1/(n−1)!nx^{n-1}/n! = x^{n-1}/(n-1)!nxn−1/n!=xn−1/(n−1)! , the preceding term, so the slopes of all terms add up to the same series. slope=0+1+2x2!+3x23!+4x34!+⋯=1+x+x22!+x33!+⋯=ex\text{slope} = 0 + 1 + \dfrac{2x}{2!} + \dfrac{3x^2}{3!} + \dfrac{4x^3}{4!} + \cdots = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots = e^xslope=0+1+2!2x+3!3x2+4!4x3+⋯=1+x+2!x2+3!x3+⋯=ex
[The slope of e^x] What does the notation dfdx\dfrac{df}{dx}dxdf for the slope (derivative) of fff mean?
dfdx≡lim⁡Δx→0ΔfΔx\boldsymbol{\dfrac{df}{dx} \equiv \displaystyle\lim_{\Delta x \to 0} \frac{\Delta f}{\Delta x}}dxdf≡Δx→0limΔxΔf : Δ\DeltaΔ denotes a change of any size, ddd an infinitesimally small change. Δx\Delta xΔx is the δ\deltaδ of the slope definition and Δf=f(x+Δx)−f(x)\Delta f = f(x + \Delta x) - f(x)Δf=f(x+Δx)−f(x) . dfdfdf and dxdxdx are both essentially zero, but their ratio is generally not.Example: d(ex)dx=ex\dfrac{d(e^x)}{dx} = e^xdxd(ex)=ex and d(xn)dx=nxn−1\dfrac{d(x^n)}{d…