# Find the 2nd order Taylor Polynomial of y(x) about x = 0, given:

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!

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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 $-1/27$.

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Thank you! I did forget the 2! – JackReacher Sep 23 '12 at 4:50

Alternative:

As a first approximation, let $y=a$ be a constant and match it to the constant term in $x$:

$$a^3+a^2+a-3\to0.$$

We have the solution $a=1$.

Then try a linear approximation and identify the first order terms:

$$(bx+1)^3+(bx+1)^2+(bx+1)-3=\cdots+3bx+2bx+bx\cdots\to x,$$giving $b=\frac16$.

Then add a second degree term, and

$$(cx^2+\frac x6+1)^3+(cx^2+\frac x6+1)^2+(cx^2+\frac x6+1)-3=\\\cdots+3cx^2+\frac{x^2}{12}+2cx^2+\frac{x^2}{18}+cx^2\cdots\to0x^2,$$or $c=-\frac1{54}$.

$$y\approx1+\frac x6-\frac{x^2}{54}\cdots.$$

This coincides with the Taylor development.

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