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Taylor's Theorem!

zkee

New member
Feb 13, 2014
2
Hey guys,

So i missed a class on Taylor's theorem and i'm getting stuck with these 2 questions. Any help would be greatly appreciated!

q1.jpgq2.jpg
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Hey guys,

So i missed a class on Taylor's theorem and i'm getting stuck with these 2 questions. Any help would be greatly appreciated!

View attachment 1964View attachment 1965
Hi zkee, and welcome to MHB!

In q.4, you write the given equation as $\dfrac{dq}q = \dfrac{3p\,dp}{p^2+1}$ ($dp$ is missing in the model solution), and integrate both sides: $$\int\dfrac{dq}q = \int\dfrac{3p\,dp}{p^2+1}.$$ Now make the substitution $z = p^2+1$, so that $dz = 2p\,dp$. That gives you the next line of the model solution, and takes you through as far as the line $\ln q = -\frac32\ln(p^2+1) + c$. Now use properties of the logarithm to get $\frac32\ln(p^2+1) = \ln(p^2+1)^{3/2}$ and $\ln q + \ln(p^2+1)^{3/2} = \ln\left[q(p^2+1)^{3/2}\right]$.

In q.5, the idea is that you are only interested in powers of $x$ up to $x^4$, so you just ignore any higher powers of $x$. So when you expand $\Bigl(x - \frac{x^2}2 + \frac{x^3}3 - \frac{x^4}4\Bigr)^2$, the only terms that survive are $x^2 - 2x\frac{x^2}2 + \frac{x^4}4 + 2x\frac{x^3}3.$