Expand (b^2 + y^2)^(3/2) as a Taylor's series In terms of y/b

In summary: I also don't understand how you could have gotten A's in all of your algebra classes as you said, unless you're just trolling.
  • #1
leo255
57
2

Homework Statement



[/B]
Expand (b^2 + y^2)^(3/2) as a Taylor's series In terms of y/b

Homework Equations

The Attempt at a Solution


[/B]
Hello, I'm clueless on this one. I tried to factor out b^3 -> b^3(1 + y^3), but can't figure out what to do next.

Thanks in advance.
 
Physics news on Phys.org
  • #2
leo255 said:

Homework Statement



[/B]
Expand (b^2 + y^2)^(3/2) as a Taylor's series In terms of y/b

Homework Equations

The Attempt at a Solution


[/B]
Hello, I'm clueless on this one. I tried to factor out b^3 -> b^3(1 + y^3)
Write b2 + y2 as b2(1 + y2/b2), and bring it out as (b2)3/2, or b3.
leo255 said:
, but can't figure out what to do next.

Thanks in advance.
 
  • #3
leo255 said:

Homework Statement



[/B]
Expand (b^2 + y^2)^(3/2) as a Taylor's series In terms of y/b

Homework Equations

The Attempt at a Solution


[/B]
Hello, I'm clueless on this one. I tried to factor out b^3 -> b^3(1 + y^3), but can't figure out what to do next.

Thanks in advance.
It looks like you did the following: ##(b^2 + y^2)^{3/2} = (b^2)^{3/2} + (y^2)^{3/2} = b^3 + y^3##. If so, don't do that. It's wrong.
 
  • #4
Sorry for the delay - This makes much more sense now. So, once you pull out the b^2, you have:

(b^2 (1 + y^2 / b^2))^3/2 = b^3 (1 + y^3 / b^3) [b^3's cancel out] = (1 + y^3) ---> p = 1, and x = y^3, ready to be substituted into Binomial Series.

Series:

1 + px + p(p-1)x / 2!... = 1 + y^3 + (y^3)(y^3 - 1) / 2! + (y^3)(y^3 - 1)(y^3 - (n - 1) / n!)
 
  • #5
You really need to review basic algebra. First off, ##(a+b)^n \ne a^n + b^n##. Even if you could do that, the ##b^3##'s wouldn't cancel out.
 
  • #6
Thanks for the reply, but I really didn't see a need to insult me (simply telling me what I did wrong should be sufficient). I am pretty good at algebra (I've gotten A's in all of my Algebra classes). The problem is that my mind is focusing on making sure I understand the problem, and I may overlook some things, like basic factorization.
 
Last edited:
  • #7
It wasn't meant as an insult; it was advice. You've made really basic algebra mistakes. You didn't factor the expression correctly in the original post; you incorrectly distributed the exponents; and you canceled terms you can't cancel. I'll note that I pointed out the mistake about the exponents in post 3, and you didn't seem to get what you did was wrong because you repeated the mistake in your subsequent post. Perhaps it was just a fluke, and you don't commonly make these types of errors. On the other hand, every step that you've written in this thread has had a major mistake in it.

It would be one thing if you were currently taking algebra, but it appears you're in calculus because you're learning about the Taylor series. At this point, you really should have these basics down.
 
  • #8
vela said:
You really need to review basic algebra.

leo255 said:
Thanks for the reply, but I really didn't see a need to insult me (simply telling me what I did wrong should be sufficient).
How is that an insult? If you're posting a calculus problem such as the one in this thread, and you make a number of conceptual errors at the algebra level, then clearly, vela's comment applies.
 
Last edited by a moderator:

What is a Taylor's series?

A Taylor's series is a mathematical representation of a function as an infinite sum of terms, where each term is a polynomial function of the variable, evaluated at a specific point. It is used to approximate a function and can be used to find values of the function at points where it is not explicitly defined.

How is a Taylor's series expanded?

A Taylor's series can be expanded by taking derivatives of the function at a specific point and evaluating them at that point. The general formula for expanding a function f(x) at the point a is: f(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)^2 + (f'''(a)/3!)(x-a)^3 + ...

What is the significance of (b^2 + y^2)^(3/2) in this Taylor's series?

The expression (b^2 + y^2)^(3/2) is the function that is being expanded as a Taylor's series. It is being expanded in terms of the variable y/b, meaning that the terms in the series will be in powers of y/b.

Why is it important to expand (b^2 + y^2)^(3/2) as a Taylor's series in terms of y/b?

Expanding (b^2 + y^2)^(3/2) as a Taylor's series in terms of y/b allows us to approximate the function for a range of values of y/b. This is useful when the function is not defined for certain values, such as when y and b approach zero. It also allows us to easily manipulate and solve equations involving the function.

What is the purpose of using a Taylor's series for (b^2 + y^2)^(3/2)?

The purpose of using a Taylor's series for (b^2 + y^2)^(3/2) is to approximate the function and calculate values of the function at points where it is not explicitly defined. This can be useful in various mathematical and scientific applications.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
679
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
998
  • Calculus and Beyond Homework Help
Replies
2
Views
182
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
711
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
Back
Top