How do you expand algebraic expressions to the nth power?

  • Thread starter Thread starter WilliamK
  • Start date Start date
  • Tags Tags
    Expressions Power
AI Thread Summary
To expand algebraic expressions to the nth power, one can use the binomial theorem, which provides a formula for expanding expressions like (x + dx)^n. The discussion highlights examples of expanding (x + dx)^2, (x + dx)^3, and (x + dx)^4, illustrating how to derive the coefficients and terms involved in each expansion. The user seeks a clear explanation and resources due to a lack of foundational knowledge in algebra and calculus. External links to Wikipedia and a resource on Pascal's triangle are provided for further clarification. Understanding these expansions is crucial for differentiating functions like y = x^n.
WilliamK
Messages
3
Reaction score
0
I would like to start by saying that I'm not at school, I'm 40 years old, and learning calculus for the first time - personal, private study - so anyone helping me out won't be giving me the answers to any course work or school work. I don't have a teacher, and there's no one I can ask who can help me, so I've come to this forum hoping some kind soul can help me.

Homework Statement



I'm trying to understand how to differentiate y=x^n, but I get stuck at the expansion stage

Homework Equations



In all cases, we are increasing y by a small amount (dy)

Example 1:
y+dy = (x+dx)^2
expanded out, it becomes: y+dy = x^2 + 2x.dx+(dx)2

Example 2:
y+dy = (x+dx)^3
expanded out, it becomes y+dy = x^3 + 3x^2.dx+3x(dx)^2+(dx)^3

Final Example:
y+dy = (x+dx)^4
expanded out, it becomes y+dy = x^4 + 4x^3dx+6x^2(dx)^2 + 4x(dx)^3 + (dx)^4

My calculus textbook assumes a knowledge I don't have.

Can someone please explain clearly & simply how one is supposed to derive these expanded expressions? Or even refer me to an external link which explains it?

Many thanks
 
Last edited:
Physics news on Phys.org
Wikipedia explains it nicely. http://en.wikipedia.org/wiki/Binomial_theorem" If you don't understand something there, feel free to ask.
 
Last edited by a moderator:
Thanks for the link, Sourabh, I'll take a look. (Its been 25 years since I last did algebra, and I'm not even sure if I covered binominal theorum at school, lol)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

Similar threads

Back
Top