Taylor's Theorem: Get Help with 2 Difficult Questions

  • MHB
  • Thread starter zkee
  • Start date
  • Tags
    Theorem
In summary, the conversation is about someone who missed a class on Taylor's theorem and is struggling with two questions. They are seeking help on the Physics Forums website. One question involves integrating an equation and using logarithm properties, while the other involves expanding a polynomial and only considering powers up to x^4.
  • #1
zkee
2
0
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
 

Attachments

  • q1.jpg
    q1.jpg
    35.3 KB · Views: 75
  • q2.jpg
    q2.jpg
    48.3 KB · Views: 78
Physics news on Phys.org
  • #2
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 $x^2 - 2x\frac{x^2}2 + \frac{x^4}4 + 2x\frac{x^3}3.$
 

FAQ: Taylor's Theorem: Get Help with 2 Difficult Questions

What is Taylor's Theorem?

Taylor's Theorem is a mathematical theorem that allows us to approximate a function using a polynomial.

How is Taylor's Theorem used?

Taylor's Theorem is used in calculus to help us approximate functions at a specific point or within a specific interval. It allows us to represent complex functions with simpler polynomial functions, making it easier to calculate values and solve problems.

What is the difference between Taylor's Theorem and Maclaurin's Theorem?

Taylor's Theorem is a generalized version of Maclaurin's Theorem. While Maclaurin's Theorem is specifically used to approximate a function at the point x=0, Taylor's Theorem can be used to approximate a function at any point within its interval.

What are the assumptions for Taylor's Theorem to hold?

The function must be differentiable and have continuous derivatives up to the desired order within the interval. The interval must also contain the point at which we want to approximate the function.

How do you find the error in Taylor's Theorem approximation?

The error in Taylor's Theorem approximation can be found by using the remainder term, also known as the Lagrange or Peano form of the remainder. This term takes into account the difference between the actual function and the polynomial approximation. The error can also be estimated by using the Lagrange error bound, which provides an upper bound for the error based on the maximum value of the n+1 derivative of the function within the interval.

Back
Top