Porisma

Worked solution

Practise a differential relation and a logarithm of secant

(a) It is given that dy/dx=y+2xdy/dx=y+2x and that y=1 when x=0. Find the first three terms in the Maclaurin series for y.

(b) Show that the Maclaurin series for ln⁡(sec⁡x)\ln(\sec x), up to and including the term in x4x^4, is 12x2+112x4\tfrac12x^2+\tfrac1{12}x^4.

CJC · Lecture · Self-Practice 2

(a) Obtain the first three coefficients

Both copies give the answer 1+x+32x21+x+\tfrac32x^2 without working. Independently, y(0)=1 and y′(0)=1y^{\prime}(0)=1. Differentiate the relation before setting x=0:

y′′=y′+2,y′′(0)=3.y^{\prime\prime}=y^{\prime}+2,\qquad y^{\prime\prime}(0)=3.

Thus y=1+1x+(3/2!)x2+⋯=1+x+32x2+⋯y=1+1x+(3/2!)x^2+\cdots=1+x+\tfrac32x^2+\cdots.

How did this part go?

(b) Use the cosine and logarithm series

The result is printed in the ask, with no separate worked answer in either copy. Independently, near zero cos⁡x>0\cos x>0 and ln⁡(sec⁡x)=−ln⁡(cos⁡x)\ln(\sec x)=-\ln(\cos x). Set u=−x2/2+x4/24+⋯u=-x^2/2+x^4/24+\cdots in the logarithm expansion:

−ln⁡(1+u)=−u+u22−⋯-\ln(1+u)=-u+\frac{u^2}{2}-\cdots

=x22−x424+12(−x22)2+⋯=x22+x412+⋯ .\begin{aligned}&=\frac{x^2}{2}-\frac{x^4}{24}\\ &{}+\frac12\left(-\frac{x^2}{2}\right)^2+\cdots\\ & =\frac{x^2}{2}+\frac{x^4}{12}+\cdots.\end{aligned}

The square of the quadratic term contributes at fourth degree and must be retained.

How did this part go?

Common mistakes

For a fourth-degree logarithm expansion, the square of an inner quadratic term is still required.

Solution notes

Both original pages compared; printed answer (a) and target expression(b) retained. All derivations here are independently supplied. No marks or attribution printed.

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