σμimplied vol · KKmax(S−K,0)udβρᵢⱼfat tailsPnL · drawdownyield · TdSₜ = μSₜ dt + σSₜ dWₜC = S·N(d₁) − K·e⁻ʳᵀ·N(d₂)σₚ² = wᵀΣwd₁ = [ln(S/K) + (r + σ²⁄2)T] ⁄ σ√T∂V/∂t + ½σ²S²·∂²V/∂S² + rS·∂V/∂S − rV = 0Sharpe = (Rₚ − R_f) ⁄ σₚVaR₉₉ = μ − 2.33σρ = Cov(X,Y) ⁄ σₓσᵧE[Rₚ] = Σ wᵢ·E[Rᵢ]φ(x) = e^(−x²/2) ⁄ √2πE[Rᵢ] = R_f + βᵢ(E[Rₘ] − R_f)dXₜ = θ(μ − Xₜ)dt + σ dWₜσ²ₜ = ω + α·ε²ₜ₋₁ + β·σ²ₜ₋₁C − P = S − K·e⁻ʳᵀΔ = ∂V/∂S Γ = ∂²V/∂S²p = (e^(rΔt) − d) ⁄ (u − d)rₜ = ln(Pₜ ⁄ Pₜ₋₁)E[Xₜ₊₁ | ℱₜ] = Xₜf* = (bp − q) ⁄ bz = (x − μ) ⁄ σP(N=k) = λᵏe^(−λ) ⁄ k!Θ = −∂V/∂Tν = S·φ(d₁)·√Tmin wᵀΣw s.t. wᵀμ = μ*∫Σ∂∞λ
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σμimplied vol · KKmax(S−K,0)udβρᵢⱼfat tailsPnL · drawdownyield · TdSₜ = μSₜ dt + σSₜ dWₜC = S·N(d₁) − K·e⁻ʳᵀ·N(d₂)σₚ² = wᵀΣwd₁ = [ln(S/K) + (r + σ²⁄2)T] ⁄ σ√T∂V/∂t + ½σ²S²·∂²V/∂S² + rS·∂V/∂S − rV = 0Sharpe = (Rₚ − R_f) ⁄ σₚVaR₉₉ = μ − 2.33σρ = Cov(X,Y) ⁄ σₓσᵧE[Rₚ] = Σ wᵢ·E[Rᵢ]φ(x) = e^(−x²/2) ⁄ √2πE[Rᵢ] = R_f + βᵢ(E[Rₘ] − R_f)dXₜ = θ(μ − Xₜ)dt + σ dWₜσ²ₜ = ω + α·ε²ₜ₋₁ + β·σ²ₜ₋₁C − P = S − K·e⁻ʳᵀΔ = ∂V/∂S Γ = ∂²V/∂S²p = (e^(rΔt) − d) ⁄ (u − d)rₜ = ln(Pₜ ⁄ Pₜ₋₁)E[Xₜ₊₁ | ℱₜ] = Xₜf* = (bp − q) ⁄ bz = (x − μ) ⁄ σP(N=k) = λᵏe^(−λ) ⁄ k!Θ = −∂V/∂Tν = S·φ(d₁)·√Tmin wᵀΣw s.t. wᵀμ = μ*∫Σ∂∞λ
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Calculus skill›Module 3 · Approximation and dynamics›Topic 3.1·~45 minIntermediate

Taylor series

Skill positionModule 3 of 4 · Topic 1 of 2
3.1 Taylor series
—Lab score— / 100Total XP—Streak—

Replacing a function by a polynomial

Near a point, a smooth function looks like a polynomial built from its derivatives there. This is the Taylor series: starting at xxx and moving a small step hhh, the function changes by its slope times hhh, plus a correction from its curvature, plus smaller corrections from higher derivatives.

f(x+h)≈f(x)+f′(x) h+12f′′(x) h2+16f′′′(x) h3+⋯f(x+h)\approx f(x)+f'(x)\,h+\tfrac{1}{2}f''(x)\,h^2+\tfrac{1}{6}f'''(x)\,h^3+\cdotsf(x+h)≈f(x)+f′(x)h+21​f′′(x)h2+61​f′′′(x)h3+⋯(1)

Four series are worth knowing near 0: ex=1+x+x22+x36+⋯e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdotsex=1+x+2x2​+6x3​+⋯, sin⁡x=x−x36+⋯\sin x=x-\frac{x^3}{6}+\cdotssinx=x−6x3​+⋯, ln⁡(1+x)=x−x22+x33−⋯\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdotsln(1+x)=x−2x2​+3x3​−⋯ and 11−x=1+x+x2+⋯\frac{1}{1-x}=1+x+x^2+\cdots1−x1​=1+x+x2+⋯. Truncating after a few terms gives a cheap approximation whose error is about the first missing term.

Finance uses this daily. A bond price PPP depends on the yield yyy; to second order ΔP≈−D P Δy+12C P (Δy)2\Delta P\approx -D\,P\,\Delta y+\tfrac{1}{2}C\,P\,(\Delta y)^2ΔP≈−DPΔy+21​CP(Δy)2, where DDD is the modified duration (the slope term) and CCC the convexity (the curvature term). The same shape appears as delta and gamma for an option: ΔV≈Δ⋅ΔS+12Γ(ΔS)2\Delta V\approx \Delta\cdot\Delta S+\tfrac{1}{2}\Gamma(\Delta S)^2ΔV≈Δ⋅ΔS+21​Γ(ΔS)2.

Two approximationsSine: at x=0.5x=0.5x=0.5, x−x36=0.5−0.020833=0.479167x-\frac{x^3}{6}=0.5-0.020833=0.479167x−6x3​=0.5−0.020833=0.479167, while the true sin⁡0.5=0.479426\sin 0.5=0.479426sin0.5=0.479426, an error of about 0.000260.000260.00026. Bond: with P=100P=100P=100, D=7D=7D=7, C=60C=60C=60 and a yield rise of 1% (Δy=0.01\Delta y=0.01Δy=0.01), ΔP≈−7+12(60)(100)(0.0001)=−7+0.3=−6.7\Delta P\approx-7+\tfrac{1}{2}(60)(100)(0.0001)=-7+0.3=-6.7ΔP≈−7+21​(60)(100)(0.0001)=−7+0.3=−6.7. The convexity term softens the loss.
Only good near the pointA truncated Taylor polynomial is an approximation near the expansion point, and its error grows quickly with the size of the step. A 1% move in yields is fine; a 5% move needs more terms or a full repricing of the bond.

Lab · what's graded

  • The series of the exponential15 pts
  • Approximating the sine15 pts
  • The size of the error15 pts
  • A bond price change20 pts
  • Which term is the curvature10 pts
  • A series by hand15 pts
  • Functions and their series10 pts
0/ 7 answered
Tasks
1Fill the blanks15 ptsNot answered

Complete the first terms of the Taylor series of exe^xex at 0.

ex=1+x+x+x+⋯\displaystyle e^x = 1 + x + \frac{x^{\htmlData{blank=a}{\phantom{\rule[-0.35em]{2.4em}{1.35em}}}}}{\htmlData{blank=b}{\phantom{\rule[-0.35em]{2.4em}{1.35em}}}} + \frac{x^{\htmlData{blank=c}{\phantom{\rule[-0.35em]{2.4em}{1.35em}}}}}{\htmlData{blank=d}{\phantom{\rule[-0.35em]{2.4em}{1.35em}}}} + \cdotsex=1+x+x​+x​+⋯
2Numeric answer15 ptsNot answered

Use sin⁡x≈x−x36\sin x\approx x-\frac{x^3}{6}sinx≈x−6x3​ to approximate sin⁡0.5\sin 0.5sin0.5 and round to 4 decimals.

sin⁡0.5=\sin 0.5 =sin0.5=accepted: ± 0.0001
3Numeric answer15 ptsNot answered

The true value is sin⁡0.5=0.479426\sin 0.5=0.479426sin0.5=0.479426. By how much does the approximation 0.4791670.4791670.479167 miss it? Give the difference.

error=\text{error} =error=accepted: ± 0.00005
4Numeric answer20 ptsNot answered

A bond has price P=100P=100P=100, modified duration D=7D=7D=7 and convexity C=60C=60C=60. Using ΔP≈−DPΔy+12CP(Δy)2\Delta P\approx -DP\Delta y+\tfrac{1}{2}CP(\Delta y)^2ΔP≈−DPΔy+21​CP(Δy)2, find ΔP\Delta PΔP when the yield rises by 1% (Δy=0.01\Delta y=0.01Δy=0.01).

ΔP=\Delta P =ΔP=accepted: ± 0.01
5Choice10 ptsNot answered

In the bond approximation, which term captures the curvature of the price-yield curve?

6Expression15 ptsNot answered

Type the Taylor polynomial of order 2 of ln⁡(1+x)\ln(1+x)ln(1+x) at 0.

T2(x)=T_2(x) =T2​(x)=
Your formula appears here
7Match10 ptsNot answered

Match each function to the start of its Taylor series at 0.

Pick the partner of each item on the left.

exe^xex
sin⁡x\sin xsinx
ln⁡(1+x)\ln(1+x)ln(1+x)
11−x\frac{1}{1-x}1−x1​
Math modeKeyboard or LaTeX inputTask 1 of 7 · Ctrl ⏎ to check

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