Solving Induction Problems: Making it Look Like You Want It To

  • Thread starter Thread starter nastygoalie89
  • Start date Start date
  • Tags Tags
    Induction
AI Thread Summary
The discussion focuses on challenges faced in solving induction problems, particularly in demonstrating the truth of P(k+1) through algebraic manipulation. The user expresses difficulty in making the problem appear as desired, despite understanding the induction process. Key equations are provided, illustrating attempts to simplify expressions involving factorials and algebraic terms. The user emphasizes the importance of establishing a base case, assuming the statement for n = k, and proving it for n = k + 1. Mastering these steps is crucial for successfully applying mathematical induction.
nastygoalie89
Messages
17
Reaction score
0

Homework Statement


I understand the process of induction and how it works, but when it wittles down I have a hard time, as my teacher says "making the problem look like you want it to." It's just algebraic stuff showing P(k+1) is true


Homework Equations


1. (4 k+1-1) + 4k+1 is equivalent to: 4(4K+1-16) over 3

2. (1 over 2k+3) x (1 over 2k+4) is equivalent to: 1 over (2k+4)!


The Attempt at a Solution


This is after I have plugged in k+1 for n and simplified as best I could.

For 2. I work the factorial on the right side and have a (2k+2) x (2k+1) remaining after the 2k+3 and 2k+4 have canceled out.
 
Physics news on Phys.org
nastygoalie89 said:

Homework Statement


I understand the process of induction and how it works, but when it wittles down I have a hard time, as my teacher says "making the problem look like you want it to." It's just algebraic stuff showing P(k+1) is true


Homework Equations


1. (4 k+1-1) + 4k+1 is equivalent to: 4(4K+1-16) over 3

2. (1 over 2k+3) x (1 over 2k+4) is equivalent to: 1 over (2k+4)!


The Attempt at a Solution


This is after I have plugged in k+1 for n and simplified as best I could.

For 2. I work the factorial on the right side and have a (2k+2) x (2k+1) remaining after the 2k+3 and 2k+4 have canceled out.
For each of these problems you need to do three things:
  1. Establish that the statement is true for some starting point (typically n = 1 - the "base case").
  2. Assume that the statement is true for n = k.
  3. Show that if the statement is true for n = k, then the statement must also be true for n = k + 1.
 
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...

Similar threads

Replies
6
Views
2K
Replies
11
Views
2K
Replies
8
Views
2K
Replies
3
Views
2K
Replies
2
Views
2K
Replies
7
Views
2K
Replies
13
Views
2K
Back
Top