Porisma

Worked solution

Transform an exponential series without differentiating again

Let y=e1/(x+1)y=e^{1/(x+1)}. Show that yy′′−(y′)2=2y2(x+1)3yy^{\prime\prime}-(y^\prime)^2=\dfrac{2y^2}{(x+1)^3}. Hence find its Maclaurin expansion through x2x^2, and deduce the expansion of e−x/(x+1)e^{-x/(x+1)} through x2x^2.

[2007/HCI/I/8]

CJC · Discussion · Question 10

Answer

e1/(x+1)=e−ex+3e2x2+O(x3)e^{1/(x+1)}=e-ex+\tfrac{3e}2x^2+O(x^3); e−x/(x+1)=1−x+32x2+O(x3)e^{-x/(x+1)}=1-x+\tfrac32x^2+O(x^3).

Differentiate the logarithm

ln⁡y=1/(x+1)\ln y=1/(x+1) gives y′/y=−1/(x+1)2y^\prime/y=-1/(x+1)^2. Differentiating gives y′′/y−(y′)2/y2=2/(x+1)3y^{\prime\prime}/y-(y^\prime)^2/y^2=2/(x+1)^3. Multiply by y2y^2 to obtain the required identity.

How did this part go?

Evaluate the first three derivatives needed

At zero, y(0)=ey(0)=e and y′(0)=−ey^\prime(0)=-e. The identity gives e y′′(0)−e2=2e2e\,y^{\prime\prime}(0)-e^2=2e^2, so y′′(0)=3ey^{\prime\prime}(0)=3e. Hence e1/(x+1)=e−ex+3e2x2+O(x3)e^{1/(x+1)}=e-ex+\tfrac{3e}2x^2+O(x^3).

How did this part go?

Use the exponent identity

−x/(x+1)=−1+1/(x+1)-x/(x+1)=-1+1/(x+1), so e−x/(x+1)=e−1e1/(x+1)=1−x+32x2+O(x3)e^{-x/(x+1)}=e^{-1}e^{1/(x+1)}=1-x+\tfrac32x^2+O(x^3).

How did this part go?

Common mistakes

Rewrite the second exponent exactly before multiplying the first series by e⁻¹.

Read the hint