Taylor's Theorem: Get Help with 2 Difficult Questions

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SUMMARY

This discussion focuses on solving two specific questions related to Taylor's Theorem. The first question involves integrating the equation $\dfrac{dq}q = \dfrac{3p\,dp}{p^2+1}$, where the correct substitution is $z = p^2+1$. The second question addresses the expansion of the expression $\Bigl(x - \frac{x^2}2 + \frac{x^3}3 - \frac{x^4}4\Bigr)^2$, emphasizing the importance of retaining only terms up to $x^4$. Both solutions highlight the application of logarithmic properties and polynomial expansion techniques.

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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
 

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zkee said:
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!

https://www.physicsforums.com/attachments/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 $x^2 - 2x\frac{x^2}2 + \frac{x^4}4 + 2x\frac{x^3}3.$
 

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