Taylor Slider

How it works

Each round picks one of four functions - eˣ, sin(x), cos(x), or ln(1+x) - a sample point x, and an order from 2, 3, or 4 (kept low enough to sum by hand). You compute the Taylor (Maclaurin, centered at 0) approximation of that order at that x and type the number. There is no live readout - you find out only after committing.

The sample points are chosen per function: for ln(1+x) they all lie inside the series' interval of convergence (-1 < x ≤ 1), such as -0.9, -0.6, -0.4, 0.5, 0.8, 0.95; the other functions use points like -1.8, -1.3, -0.8, 0.6, 1.1, 1.5, 1.9. Each round also carries an error threshold of 0.1, 0.01, or 0.001 used in the reveal.

After checking, the reveal shows the exact order-n sum, the true function value, the resulting error, and whether your assigned order clears the round's error threshold - including the smallest order (up to 8) that does, or a note that even order 8 cannot reach it at that x.

How scoring works

Your answer is graded within a tolerance of max(0.01, 0.5% of the true partial sum). A correct answer adds 1 to score; every check adds 1 to rounds. The threshold comparison in the reveal is informational - only the typed sum is graded.

Know the four series cold

Everything in this game comes from four Maclaurin expansions. eˣ = 1 + x + x²/2 + x³/6 + x⁴/24 + …, every power present. sin(x) = x - x³/6 + x⁵/120 - …, odd powers only, signs alternating. cos(x) = 1 - x²/2 + x⁴/24 - …, even powers only, alternating. ln(1+x) = x - x²/2 + x³/3 - x⁴/4 + …, every power from 1, alternating, with 1/k denominators rather than factorials.

The parity structure matters for what 'order n' contains: an order-4 sin polynomial is just x - x³/6, because the x⁴ term of sine is zero. Likewise order-3 cosine is 1 - x²/2. When the game says order 4, sum every term of degree at most 4 that actually exists for that function - do not invent even terms for sine or odd terms for cosine.

The two 1/k-versus-1/k! families behave very differently: factorial denominators crush terms fast, so eˣ, sin, and cos converge everywhere and quickly; the harmonic denominators of ln(1+x) shrink slowly, which is why its convergence is confined to -1 < x ≤ 1 and why the game samples it only inside that interval - outside it, adding terms never helps.

Organize the arithmetic term by term

Build a small table: powers of x first, then divide by the right denominator, then apply signs, then add in one pass. Powers chain cheaply - compute x², multiply by x for x³, again for x⁴. At x = 0.6: x² = 0.36, x³ = 0.216, x⁴ = 0.1296. Keep one more decimal than you think you need; the grading tolerance is 0.5% of the sum with a floor of 0.01, so sloppy rounding across four terms can cost the round.

For alternating series (everything here except eˣ, plus ln at positive x and eˣ at negative x behaves similarly), add terms in order and watch the partial sums bracket the answer - each successive partial sum overshoots the truth in alternating directions. That bracketing is a built-in arithmetic check: if your partial sums are not alternating around a shrinking interval, you dropped a sign.

Negative x deserves extra care with odd powers: at x = -0.8, x³ = -0.512, so the ln series term -x²/2 + x³/3 becomes -0.32 - 0.1707, both negative - for negative arguments the ln series terms all push the same (negative) direction and the sum dives, which is why ln(1 - 0.9) needs many terms.

Read the error like Lagrange would

The error of a truncated Taylor series is governed by the first term you dropped: for these series the remainder after order n is roughly the size of the next nonzero term (exactly bounded by it, for alternating series with decreasing terms). So before the reveal you can estimate your own error: order-3 eˣ at x = 1.1 drops x⁴/24 ≈ 0.061 first - so expect an error near 0.06, clearing a 0.1 threshold but not 0.01.

This next-term heuristic explains the reveal's smallest-order line. To clear a threshold ε at a point x, find the first k where the term magnitude drops below ε. Factorial denominators mean each extra order multiplies the term by x/(k+1), so once k exceeds |x| the terms collapse rapidly - for |x| < 2 a few extra orders buy orders of magnitude of accuracy.

For ln(1+x) near the edge of convergence (x = 0.95 or -0.9), terms shrink like xᵏ/k - painfully slowly. That is the deep lesson the game's design encodes: convergence radius and convergence speed are different things, and an approximation that is excellent at x = 0.5 can be nearly useless at x = 0.95 at any hand-computable order.

A worked example

Round: compute the order-3 Taylor approximation of eˣ at x = 0.6.

Step 1 - list the terms. eˣ order 3: 1 + x + x²/2 + x³/6.

Step 2 - powers. x = 0.6, x² = 0.36, x³ = 0.216.

Step 3 - divide and sum. 1 + 0.6 + 0.18 + 0.036 = 1.816. Type 1.816. Tolerance is 0.5% of 1.816 ≈ 0.009 (above the 0.01 floor, so effectively about ±0.01) - 1.816 is safely graded correct.

Step 4 - anticipate the reveal. True value e^0.6 ≈ 1.82212, so the error is about 0.0061. First dropped term: x⁴/24 = 0.1296/24 = 0.0054 - the next-term heuristic predicted the error almost exactly. Against a threshold of 0.01 the order-3 sum already clears; against 0.001 it does not, and the next term size suggests order 4 (error ≈ x⁵/120 ≈ 0.00065) is the smallest that does.

Common mistakes

• Including terms that do not exist: sine has no even-power terms and cosine no odd ones. Order 4 for sin(x) means x - x³/6, nothing more.

• Using factorial denominators for ln(1+x). Its series divides by k, not k!: x - x²/2 + x³/3 - x⁴/4. Writing x³/6 there is the single most common slip.

• Dropping a sign on alternating series, especially with negative x where the parity of the power flips signs again. Track sign and power separately.

• Answering the true function value instead of the partial sum. The game grades the order-n sum; at low orders and larger x, the two differ by more than the tolerance and the calculator answer is marked wrong.

• Rounding each term to 2 decimals before summing. Four rounded terms can drift beyond the 0.5% tolerance - carry 4 or 5 decimals through and round once at the end.

Why interviews test this

Taylor expansion is arguably the most-used tool in quant interviews: quick numeric estimates (e^0.1, ln(1.05)), the log-return approximation ln(1+x) ≈ x - x²/2 behind volatility drag, Itô's lemma as a second-order expansion, delta-gamma P&L approximations, and duration-convexity in rates are all order-2 Taylor arguments. Being able to sum to order 3 or 4 by hand, cleanly, is table stakes.

The error question is the differentiator: interviewers follow up with how big is the error and when does the expansion fail. The next-dropped-term estimate and the convergence-interval caveat for ln(1+x) - the exact two facts this game's reveal drills - are the expected answers.