Find Coeff. of x^{88} & Solve for x

  • Thread starter Thread starter mercedesbenz
  • Start date Start date
AI Thread Summary
The discussion focuses on finding the coefficient of x^{88} in the polynomial expression (x+1)(x+2)(x-3)...(x+89)(x-90) and solving for x in the equation 108/√(x^2-2916) = (378-x-√(x^2-2916))/(x+54). A hint is provided to first determine the coefficients of x^{90} and x^{89} to aid in finding the desired coefficient. For the second problem, participants are encouraged to simplify the equation by calculating √2916 and the value of 378/√2916. The original poster confirms that they have successfully solved the problems.
mercedesbenz
Messages
14
Reaction score
0
[SOLVED] please hint me

Homework Statement


28. What is the coefficient of x^{88} from
(x+1)(x+2)(x-3)(x+4)(x+5)(x-6)...(x+88)(x+89)(x-90).

29. Find x from this equation \frac{108}{\sqrt{x^2-2916}}=\frac{378-x-\sqrt{x^2-2916}}{x+54}.


Homework Equations





The Attempt at a Solution


 
Physics news on Phys.org
mercedesbenz said:
28. What is the coefficient of x^{88} from
(x+1)(x+2)(x-3)(x+4)(x+5)(x-6)...(x+88)(x+89)(x-90).

Hi mercedesbenz! :smile:

ok … a hint, as you asked for …

First, try answering: what are the coefficients of x^{90} and of x^{89} ? :smile:
29. Find x from this equation \frac{108}{\sqrt{x^2-2916}}=\frac{378-x-\sqrt{x^2-2916}}{x+54}.

You can simplify this slightly, before solving it:

Hint: what is √2916? What is 378/√2916? :smile:
 
Thank you so much I've already solved it.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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...
Back
Top