Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am given the equation:

$$y^3 + y^2 + y - 3 = x$$

with $y(0)=1$. I am wondering if what I have done is valid, given that this is a homework question for an Integral Calculus class, but I seem to be able to answer the question without doing any integration.

Taking the derivative of both sides (and using implicit differentiation) with respect to $x$ and then rearranging, I have

$$\frac{d}{dx} \left(y^3 + y^2 + y - 3\right) = \frac{d}{dx}(x)$$

$$\frac{dy}{dx} = \frac{1}{3y^2 + 2y + 1}$$

So, since I am given $y(0) = 1$, I have my first two terms in the Taylor Series: f(0) = 1, and now f'(0) = 1/6, since y = 1 when x = 0.

I then take the 2nd derivative with respect to x (and again using implicit differentiation) to get:

$d^2y/dx^2 = - y'(x)(6y+2)/(3y^2 + 2y + 1)^2$

Plugging in for y and y', when x = 0, and I get y''(0) = -1/27

Therefore my Taylor polynomial of order 2, about x = 0, is:

$1 + x/6 - x^2/27$

Is this correct?

I appreciate any tips and advice!

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Minor error, recall that we need to divide by $2!$ at $x^2$ (and so on). The Taylor expansion about $x=a$ up to the $(x-a)^n$ term goes like this: $$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+ \cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n.$$

You got after differentiating once, $$(3y^2+2y+1)\frac{dy}{dx}=1.$$ To find the second derivative, I prefer not to divide, it sometimes makes a mess. The above expression looks nice, might as well just differentiate as it stands. We get $$(3y^2+2y+1)\frac{d^2y}{dx^2}+ (6y+2)\left(\frac{dy}{dx}\right)^2=0,$$ and now we can find the second derivative at $0$. It does indeed simplify to $-1/27$.

share|improve this answer
    
Thank you! I did forget the 2! –  JackReacher Sep 23 '12 at 4:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.