7:02 PM. The first problem on the sheet is a spring. A mass on it is pulled down and let go, the friction eats the motion, and the displacement follows x(t) = e−2t sin(5t). The question wants the Laplace transform, the poles and one number: F(s) at s = 1. By 10 PM the same sheet will have asked for a product of 23 fractions and for the sum of two infinite series, one of which refuses to have a sum at all.
What follows is that evening, block by block, as the clock moved. Every number was computed in the calculators linked along the way, and the three screenshots show the exact inputs, so any line can be checked in a few seconds.
7:02 PM: a spring that stops bouncing
The textbook route goes through the table. The sine pair says L{sin(5t)} = 5 / (s2 + 25), and the factor e−2t shifts every s to s + 2. So the damped spring becomes 5 / ((s + 2)2 + 25), valid for Re(s) > −2. Evaluated at s = 1 it gives 5 / (9 + 25) = 0.147059.
The poles are where the denominator vanishes: (s + 2)2 = −25, so s = −2 ± 5j. That pair is the whole story of the spring in two numbers. The real part, −2, is the rate at which the swing dies away. The imaginary part, 5, is the angular frequency, which is 0.795775 Hz, one full swing every 1.25664 s. Poles to the left of the imaginary axis mean the motion settles; drop the friction and they move onto the axis, where the plain sin(5t) would ring for ever.
7:31 PM: the same exponent, a different shape
Problem 2 keeps the e−2t and swaps the sine for t3. The power rule gives L{t3} = 3! / s4, and the same shift turns it into 6 / (s + 2)4. This time there is one pole, at s = −2, but of order 4. A repeated real pole is how a transform says "a power of t times an exponential": the function rises, peaks, and then the exponential wins. At s = 1 the value is 6/81 = 0.0740741.
7:48 PM: a switch thrown at t = 2
Problem 3 is a circuit with a switch that closes two seconds in. The input is the unit step u(t-2), and the delay turns into a factor: F(s) = e−2s / s. At s = 1 that is e−2 = 0.135335. Piecewise inputs, the kind that switch on, off and on again, are built from steps like this one, each carrying its own e−as.
| f(t), t ≥ 0 | F(s) | Converges for | Poles |
|---|---|---|---|
| e−2t sin(5t) | 5 / ((s + 2)2 + 25) | Re(s) > −2 | −2 ± 5j |
| sin(5t) | 5 / (s2 + 25) | Re(s) > 0 | ±5j |
| cos(5t) | s / (s2 + 25) | Re(s) > 0 | ±5j |
| t3 | 6 / s4 | Re(s) > 0 | 0, order 4 |
| t3 e−2t | 6 / (s + 2)4 | Re(s) > −2 | −2, order 4 |
| e−2t | 1 / (s + 2) | Re(s) > −2 | −2 |
| u(t-2) | e−2s / s | Re(s) > 0 | 0 |
Read the table from right to left and it becomes the tool for the second half of every Laplace problem: once the algebra in s is done, each piece of the answer is matched to a row and turned back into a function of time.
8:05 PM: 23 birthdays as one long product
The second block of the sheet leaves springs behind. How likely is it that 23 people in a room all have different birthdays? Line them up. The first can have any of 365 days, the second any of the remaining 364, and so on down to the 23rd, who has 343 free days. The probability is a product of 23 fractions:
(365/365) × (364/365) × (363/365) × ... × (343/365)
In capital pi notation that is the product of (366 − n)/365 for n from 1 to 23, and it comes out at 0.492703. Just under a half, which makes a shared birthday slightly more likely than not in a group this small.
The running product in the table under the result shows the moment it happens. After 22 people it is still 0.524305; the 23rd person pushes it below one half. What surprises people is not the product itself but how slowly each factor shrinks and how quickly they add up: every fraction is above 0.93, yet 23 of them together cut the chance in half.
| People in the room | All birthdays different | At least one shared |
|---|---|---|
| 10 | 0.883052 | 0.116948 |
| 20 | 0.588562 | 0.411438 |
| 22 | 0.524305 | 0.475695 |
| 23 | 0.492703 | 0.507297 |
| 30 | 0.293684 | 0.706316 |
| 40 | 0.108768 | 0.891232 |
| 50 | 0.0296264 | 0.970374 |
| 57 | 0.00987754 | 0.990122 |
By 57 people a shared birthday is a near certainty at 0.990122. The model assumes 365 equally likely days and ignores leap days and twins, which is why real classrooms match slightly more often than the table says.
8:30 PM: odd numbers and a unit digit
The next line on the sheet looks innocent: multiply the first ten odd numbers. It is an arithmetic sequence with first term 1 and difference 2, so the same calculator handles it. 1 × 3 × 5 × ... × 19 = 654,729,075. The follow-up asks for the unit digit of the product of the first 50 odd numbers. That product is about 2.72539 × 1078, a 79-digit number, and the calculator lists every digit. The last one is 5, and it has to be: the product contains the factor 5 and no factor 2, so it ends in 5 whatever the length.
The even version grows faster because each term is one larger. The first ten even numbers multiply to 210 × 10! = 3,715,891,200, since factoring a 2 out of every term leaves 1 × 2 × ... × 10.
8:47 PM: a product that creeps up on pi/2
The last product of the block is a famous one. Multiply (4n2) / (4n2 − 1) for n = 1, 2, 3, and so on: 4/3 × 16/15 × 36/35 × ... Ten factors give 1.53385. A thousand give 1.5704. The target is π/2 = 1.5708, the Wallis product, and it gets there with painful slowness: after a thousand factors the fourth decimal is still wrong. That slowness is the bridge to the last hour.
9:03 PM: the Basel problem, and what a hundred terms buy
The third block is about sums that go on for ever. The first is 1 + 1/4 + 1/9 + 1/16 + ..., the reciprocals of the squares. The p-series test settles the question of convergence in one line: the power is 2, which is greater than 1, so the sum is finite. The value is π2/6 = 1.644934067, the answer to the Basel problem.
The interesting number is the gap. A hundred terms add up to 1.6349839, still 0.00995 short of the limit. The remainder after N terms is close to 1/N, so each extra correct digit costs ten times as many terms: a guaranteed error below 0.001 needs 1,001 terms. Convergent is not the same as fast.
Raise the power and the picture changes. The sum of 1/n3 is 1.202056903, a constant with no known closed form, and its tail shrinks like 1/N2, so it is already correct to about two decimals after ten terms.
9:26 PM: a sum that never ends
Then the sheet drops the power to 1, and everything breaks. The harmonic series 1 + 1/2 + 1/3 + ... has terms that shrink to zero, yet it has no sum. Its partial sums grow without limit, only very slowly. The first 100 terms add up to 5.187377518. The first 1,000 give 7.485470861. To pass 10 takes 12,367 terms: after 12,366 the sum is 9.999962148, and the next term lifts it to 10.00004301.
The partial sums track ln N + 0.5772, so each extra unit of height costs about 2.7 times as many terms as the last one, and the sum still passes every number eventually. This is the standard lesson of the evening: terms going to zero is necessary for a series to converge, never enough. With a power of one half, the sum of 1/√n diverges even faster, reaching 18.58960382 after only 100 terms.
9:44 PM: flip every other sign
The last problem takes the divergent harmonic series and alternates the signs: 1 − 1/2 + 1/3 − 1/4 + ... Now it converges, to ln 2 = 0.6931471806. The alternating series test does the work: the terms shrink steadily to zero and the signs take turns, so every partial sum overshoots or undershoots the limit by less than the next term. Ten terms give 0.6456349206. Three correct decimals, guaranteed, take 1,000 terms.
The convergence is only conditional, because the same terms without the minus signs are the harmonic series again. That fragility has a strange consequence: reorder the terms and the sum can be made to come out as any number at all.
| Series | Error < 0.01 | Error < 0.001 | Error < 0.0001 |
|---|---|---|---|
| 1 + 1/4 + 1/9 + ... (p = 2) | 101 | 1,001 | 10,001 |
| 1 - 1/2 + 1/3 - ... (alternating harmonic) | 100 | 1,000 | 10,000 |
| 1/(1·2) + 1/(2·3) + ... (telescoping) | 100 | 1,000 | 10,000 |
| 1 + 1/2 + 1/4 + ... (geometric, r = 0.5) | 8 | 11 | 15 |
Terms needed to guarantee each accuracy, as the series calculator reports them. Three of the four pay ten times more for every extra digit. The geometric series pays three or four terms per digit, because each term halves the gap.
10:00 PM: what the evening added up to
Three blocks, one idea in different clothes. A Laplace transform exists only where an integral to infinity converges, which is why every pair on the sheet came with a region of convergence. A product of shrinking fractions tells the birthday story, and one of infinitely many factors creeps up on π/2. A sum of shrinking terms may settle, as 1/n2 does on π2/6, or wander off for ever, as 1/n does, and the difference between those two fates is a single step in the exponent. Before trusting any infinite process, ask whether it converges, and then ask how fast. The first question decides whether there is an answer at all; the second decides whether a hundred terms, or a thousand, are anywhere near it.
Tools discussed in this article
- Laplace transform calculator: F(s) for powers of t (including fractional ones through the Gamma function), exponentials, sinh and cosh, sine and cosine, damped oscillations, the unit step, the Dirac delta and constants, with the region of convergence, the poles and the value of F(s) at any real s.
- Product notation calculator: capital pi over an arithmetic sequence, a geometric sequence or any formula in n, kept exact as whole numbers and fractions, with closed forms and the running product after every term.
- Series convergence calculator: geometric, p-series, alternating p-series and telescoping series, with the deciding test, the exact sum, partial sums of up to a million terms and the number of terms any accuracy needs.
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