Final Thoughts — Question 2

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Question 2

The exact solution of y′=yy'=y, y(0)=1y(0)=1, is supplied as y=exy=e^x. A polynomial approximation is P(x)=1+x+x22+x36.P(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}. You may use Taylor’s theorem with a fourth-derivative remainder.

Tasks

  1. Calculate the residual P′−PP'-P and determine whether PP solves the IVP exactly on any open interval containing 00.

  2. Let E(x)=ex−P(x)E(x)=e^x-P(x). Prove that for 0≤x≤10\le x\le 1, x424≤E(x)≤ex424.\frac{x^4}{24}\le E(x)\le\frac{e x^4}{24}.

  3. Give exact lower and upper bounds for E(1)E(1), and identify whether PP overestimates or underestimates the solution for 0<x≤10<x\le 1.

  4. Plot the error and its two bounds in your solution. Explain why matching the initial value and several derivatives at 00 does not make the polynomial an exact interval solution.

Original worksheet page 1: question and worked solution for 1-3-002
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Question 2 – Solution

Strategy. Use a residual to test exactness and Taylor’s theorem to assess the approximation. These answer different questions.

Step 1: Exactness test. Since P′=1+x+x2/2P'=1+x+x^2/2, P′−P=−x36.\boxed{P'-P=-\frac{x^3}{6}.} The residual is nonzero whenever x≠0x\ne 0. Although P(0)=1P(0)=1, the polynomial does not solve the equation on any open interval containing 00.

Step 2: Error bounds. For 0<x≤10<x\le 1, Taylor’s theorem supplies a number ξ\xi with 0<ξ<x0<\xi<x such that E(x)=eξx424.E(x)=\frac{e^\xi x^4}{24}. Because 1≤eξ≤e1\le e^\xi\le e, the requested bounds follow; at x=0x=0 all three expressions are zero.

See the diagram in the original worksheet below.

Step 3: Size and sign. At x=1x=1, P(1)=8/3P(1)=8/3, so 124≤E(1)=e−83≤e24.\boxed{\frac 1{24}\le E(1)=e-\frac 83\le\frac e{24}.} For x>0x>0 the lower bound is positive, so PP underestimates the exact solution. The two endpoint bounds are in fact strict at 11.

Step 4: Local agreement is limited evidence. The values of PP and its first three derivatives agree with those of exe^x at 00. That is local Taylor information. Solving the ODE requires P′(x)=P(x)P'(x)=P(x) at every point of an interval, which the nonzero residual disproves.

Original worksheet page 2: question and worked solution for 1-3-002

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