Real Solutions of 4th Degree Polynomial Equation

[Raghav Gupta]

Homework Statement

To find number of real solutions of:
$\frac{1}{x-1}$ $+\frac{1}{x-2}$ + $\frac{1}{x-3}$ + $\frac{1}{x-4}$ =2[/B]

Homework Equations

It will form a 4th degree polynomial equation.

The Attempt at a Solution

The real solutions could be 0 or 2 or 4 as complex solutions always exist in pairs.
It is a 4th degree polynomial equation , the real roots could be 4 but how?[/B]

 

[Stephen Tashi]

The given equation is not a 4th degree polynomial equation. When you do multiplications on it to form a 4th degree polynomial equation, you may create an equation that has a different solution set than the original equation. For example, the original equation does not have solutions x = 1, x = 2, x =3 or x = 4 since such values require division by zero. (For example, for \frac {1}{x-2} to have a defined value, we cannot have x = 2.

Are you studying functions that have asymptotes? Perhaps you are expected to sketch a graph of the function f(x) = \frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3} + \frac{1}{x-4} - 2. The question asked for the number of roots ( the number of places the graph of the function crosses the x-axis) and not the particular values of the roots.

 

[Raghav Gupta]

Sketching graph of that function seems difficult. I am not studying functions that have asymptotes. I only require number of real roots. Calculus can be used here I think but not able to apply. I know that the number of real roots maybe 0 or 2 or 4.

 

[BvU]

If you find the combined sketch difficult, start with the first term. Then the second, then estimate what 1st + 2nd looks like. You will see the pattern quickly -- and have overcome your aversion to things that seem difficult as well :)

 

[Raghav Gupta]

If you find the combined sketch difficult, start with the first term. Then the second, then estimate what 1st + 2nd looks like. You will see the pattern quickly -- and have overcome your aversion to things that seem difficult as well :)

I was trying and found out that whenever value of x is negative y is always negative. So, there is no crossing of x-axis when x -ve, so no real root till that point. Now the problem arises that around x=5 y value is coming +ve. So there is a crossing but there may be more. Can anyone help in that?

 

[Stephen Tashi]

. I know that the number of real roots maybe 0 or 2 or 4.

As I said before, you don't know that.

Let h be a small positive number, say h = 1/100,000. What is the approxomate value of the function at x = 1 + h ?

At that value, all terms except \frac{1}{x-1} are of moderate size. The term \frac{1}{x-1} has value 100,000. Now consider the value of the function at x = 2 - h.

 

[Raghav Gupta]

As I said before, you don't know that.

Let h be a small positive number, say h = 1/100,000. What is the approxomate value of the function at x = 1 + h ?

At that value, all terms except \frac{1}{x-1} are of moderate size. The term \frac{1}{x-1} has value 100,000. Now consider the value of the function at x = 2 - h.

At x = 2 - h , the term \frac{1}{x-2} would be -100,000. It means the graph crossed x-axis one time there.
Yeah , it seems I got it a bit.
Then, at x = 2 + h the value again becomes 100,000 second crossing
Then at 3-h , -100,000 third crossing
At 3+h fourth crossing,
4-h fifth crossing
4+h sixth crossing
Then at value between 5 and 6 seventh crossing and after that no crossing.
But from this the answer is coming 7.
I know the answer is 4 from calculators.
So, what is my mistake?

 

[Stephen Tashi]

At x = 2 - h , the term \frac{1}{x-2} would be -100,000. It means the graph crossed x-axis one time there.

If you know the function was positive for some x < 2 then you could say there has been a crossing. But was f(x) ever positive for x < 2 ?

Yeah , it seems I got it a bit.
Then, at x = 2 + h the value again becomes 100,000 second crossing

You have the basic idea. But to determine a crossing you must consider what happens between two places where f(x) "blows up", not just "at" one place.

Considering x = 1 + h and x = 2-h is what tells you that the graph crosses the x-axis between x = 1 and x = 2.

 

[Ray Vickson]

At x = 2 - h , the term \frac{1}{x-2} would be -100,000. It means the graph crossed x-axis one time there.
Yeah , it seems I got it a bit.
Then, at x = 2 + h the value again becomes 100,000 second crossing
Then at 3-h , -100,000 third crossing
At 3+h fourth crossing,
4-h fifth crossing
4+h sixth crossing
Then at value between 5 and 6 seventh crossing and after that no crossing.
But from this the answer is coming 7.
I know the answer is 4 from calculators.
So, what is my mistake?

I think you are mis-understanding what is meant by "crossing". When the graph of y = f(x) jumps from y = -∞ to y = +∞ as x passes from left to right through x = 1 (or x = 2, 3, or 4) we do not regard that as a "crossing", because there is no value of x in the interval (1-h,1+h) that gives f(x) = 0 exactly. Of course, if you draw the graph of y = f (x) and put in a vertical line at x = 1 (or x = 2, etc.) that vertical line does pass through the x-axis; but the line is not actually part of the graph.

Anyway, x-axis crossings are not the issue in this problem: y = 2 crossings are what matter.

 

[LCKurtz]

@Ray Vickson

Hey Ray, have you ever seen this with Maple? I tried to plot the given function and the line y=2 and got this output:

10 + BB BB BB B B
+ BB BB BB B B
8 + BB BB BB B B
+ BB BB B B B B
6 + BB BB B B B B
+ BB B B B B B BB
4 + BB B B B BB B BB
+ B B B BB B B B BB
AAAAAAAAA2A*AAAAAAAAA*A*AAAAAAA*AA*AAAAAA*AA**AAAAA*AAAAAA******AAAAAAAAA
+ B BB B BB B BB B BBBBBBBBB
-+-+-+-+--+-+-+-+-+-+-*-+**-+-+-*-+-**+-+-*-+-+**-+-*-+-+-+-+-+-+-+--+-+-+-
+ B BB B B B BB B
-1BBBBBBBB + 1 BB 2 BB 3 BB 4 5 6
-2B*BBB B B B B B B B
+ BB B B B B B B B
-4 + BB B B B B B BB
+ B B B B B B BB
-6 + BB B B B BB BB
+ B B B B BB BB
-8 + B B BB BB BB
+ B B BB BB BB
-10 + B B BB BB BB


It's obviously some sort of ASCII output in a very rough shape of the graph. I'm using Maple 13.02 but I don't know how it got in this mode or how to get it out of it.

 

[Raghav Gupta]

Anyway, x-axis crossings are not the issue in this problem: y = 2 crossings are what matter.

Sorry Ray as I am not able to get it. Can you please elaborate.

If you know the function was positive for some x < 2 then you could say there has been a crossing. But was f(x) ever positive for x < 2 ? You have the basic idea. But to determine a crossing you must consider what happens between two places where f(x) "blows up", not just "at" one place.

Considering x = 1 + h and x = 2-h is what tells you that the graph crosses the x-axis between x = 1 and x = 2.

It is seeming a hard puzzle to me what you said Stephen. Isn't it true that for x= 1 + h that is x < 2 f(x) is positive?

 

[Stephen Tashi]

It is seeming a hard puzzle to me what you said Stephen. Isn't it true that for x= 1 + h that is x < 2 f(x) is positive?

I didn't mean to imply that f(x) was never positive for x < 2. I meant that you must check that x was positive for some x < 2 before you conclude that there was a place where x = 0 between that place and x = 2 - h.

An example of what Ray is saying is that x = 2 is not a place where the graph crosses the x-axis, even though f(2-h) is negative and f(2+h) is positive. The value at x = 2 is a "singularity". It is a place where the graph does not exist.

 

[Raghav Gupta]

An example of what Ray is saying is that x = 2 is not a place where the graph crosses the x-axis, even though f(2-h) is negative and f(2+h) is positive. The value at x = 2 is a "singularity". It is a place where the graph does not exist.

So, how now to go for it?

 

[Raghav Gupta]

See this
http://m.wolframalpha.com/input/?i=graph of 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4)-2&x=0&y=0
This is showing I think no solutions.
But see also this
http://m.wolframalpha.com/input/?i=solution+of+1/(x-1)+++1/(x-2)+++1/(x-3)+++1/(x-4)-2&x=0&y=0
Showing 4 solutions?

 

[Stephen Tashi]

So, how now to go for it?

For example, think about what happens between x = 1+h and x = 2 -h instead of thinking of something between x = 2-h and x = 2 + h.

After that think about what happens "at the ends". What happens between when x is very small and x = 1-h? What happens between x = 4+h and when x becomes very large.

 

[Raghav Gupta]

Please can you refer my above post?
For my better understanding with diagram.

 

[Ray Vickson]

Hey Ray, have you ever seen this with Maple? I tried to plot the given function and the line y=2 and got this output:

10 + BB BB BB B B
+ BB BB BB B B
8 + BB BB BB B B
+ BB BB B B B B
6 + BB BB B B B B
+ BB B B B B B BB
4 + BB B B B BB B BB
+ B B B BB B B B BB
AAAAAAAAA2A*AAAAAAAAA*A*AAAAAAA*AA*AAAAAA*AA**AAAAA*AAAAAA******AAAAAAAAA
+ B BB B BB B BB B BBBBBBBBB
-+-+-+-+--+-+-+-+-+-+-*-+**-+-+-*-+-**+-+-*-+-+**-+-*-+-+-+-+-+-+-+--+-+-+-
+ B BB B B B BB B
-1BBBBBBBB + 1 BB 2 BB 3 BB 4 5 6
-2B*BBB B B B B B B B
+ BB B B B B B B B
-4 + BB B B B B B BB
+ B B B B B B BB
-6 + BB B B B BB BB
+ B B B B BB BB
-8 + B B BB BB BB
+ B B BB BB BB
-10 + B B BB BB BB


It's obviously some sort of ASCII output in a very rough shape of the graph. I'm using Maple 13.02 but I don't know how it got in this mode or how to get it out of it.

No, I have never seen it, and when I give the following instructions (in Maple 11 on an HP Probook) everything works just fine---it even puts in the vertical line:
f:=1/(x-1):
pflot(f,x=-1..3,y=-10..10);

To plot it without the vertical line I just said
plot(f,x=-1..3, y=-10..10,discont=true);

Similarly,
g = add(1/(x-i),i=1..4): plot(g, x=-1..6, y=-10..10);
produces a graph with 4 vertical lines at x = 1,2,3,4.

 

[Stephen Tashi]

Please can you refer my above post?
For my better understanding with diagram.

The plot in your link isn't an informative plot. It doesn't clearly what happens between x = 0 and x = 5 and that interval is the important part of the graph. (I don't use Wolfram, so I can't advise you how to get a clearer plot. Can you plot only the interval between x = 0 and x = 5? Perhaps it will show up better that way.)

 

[Ray Vickson]

Please can you refer my above post?
For my better understanding with diagram.

If you want to understand what is happening, forget about trying to use Wolfram Alpha, and just make simple sketches done manually. (Yes, I am serious!) For $f(x) = 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4)$, you can understand what happens near $x = 1$ just by looking at the simple function $1/(x-1)$. Slightly to the left of $x = 1$ this function is large negative and slightly to the right it is large positive. Basically, it jumps from $y = -\infty$ to $y = +\infty$ as $x$ passes through 1, from left to right. The other terms $1/(x-2) + 1/(x-3) + 1/(x-4)$ will just change things a bit near $x = 1$ but will not alter in any way the jump behavior at $x = 1$. (You should thing about why that is true!)

Similarly, you can understand what happens near $x = 2, 3, 4$ just by looking at the simpler functions $1/(x-2), \; 1/(x-3), \; 1/(x-4)$.

So, just to the right of $x = 1$ the function is large positive, and just to the left of $x = 2$ it is large negative, which means that between $x=1$ and $x=2$ there is a root of your equation $f(x) = 2$. Furthermore, you should think about why there is just one, single root in the interval $(1,2)$.

Keep going like that for other $x$-intervals.

 

[SammyS]

Please can you refer my above post?
For my better understanding with diagram.

The plot in your link isn't an informative plot. It doesn't clearly what happens between x = 0 and x = 5 and that interval is the important part of the graph. (I don't use Wolfram, so I can't advise you how to get a clearer plot. Can you plot only the interval between x = 0 and x = 5? Perhaps it will show up better that way.)

Try this in Wolfram Alpha.

http://m.wolframalpha.com/input/?i=...1/(x-3)+++1/(x-4)-2,+from+x=0+to+x=7&x=4&y=10

That takes x from 0 to 7.

If you simply plot $\ f(x) = 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4)\$ without subtracting 2, you may get some idea regarding the methods Steven and Ray are suggesting.

 

[LCKurtz]

@Ray Vickson

No, I have never seen it, and when I give the following instructions (in Maple 11 on an HP Probook) everything works just fine---it even puts in the vertical line:
f:=1/(x-1):
pflot(f,x=-1..3,y=-10..10);

.

Here's what's on my worksheet right now:
(Actually it puts spaces so the A's look like a parabola). I don't know where the smiley came from.
restart:with(plots):
plot(x^2,x = -2..2);
Here's the output:

A                                  4 +                                    A 
  AA                                   +                                   AA 
    A                                  +                                  A  
    AA                                 +                                 AA  
     AA                                +                                A    
       A                             3 +                               A      
       AA                              +                              AA      
         A                             +                             A        
          A                            +                            A        
          AA                           +                           AA        
            AA                         |                         AA          
             AA                      2 +                        AA            
              AA                       +                       AA            
                AA                     +                     AA              
                 AA                    +                    AA                
                  AAA                  +                  AAA                
                    AAA              1 +                 A                    
                      AAA              +              AAA                    
                        AAA            +            AAA                      
                           AAA         +         AAA                          
                             AAAAA     +     AAAAA                            
  +---+--+---+---+--+---+---+---+-***********-+---+---+---+--+---+---+--+---+ 
-2                -1                                     1                 2

 

[Ray Vickson]

@Ray Vickson


Here's what's on my worksheet right now:
(Actually it puts spaces so the A's look like a parabola). I don't know where the smiley came from.
restart:with(plots):
plot(x^2,x = -2..2);
Here's the output:

A                                  4 +                                    A 
  AA                                   +                                   AA 
    A                                  +                                  A  
    AA                                 +                                 AA  
     AA                                +                                A    
       A                             3 +                               A      
       AA                              +                              AA      
         A                             +                             A        
          A                            +                            A        
          AA                           +                           AA        
            AA                         |                         AA          
             AA                      2 +                        AA            
              AA                       +                       AA            
                AA                     +                     AA              
                 AA                    +                    AA                
                  AAA                  +                  AAA                
                    AAA              1 +                 A                    
                      AAA              +              AAA                    
                        AAA            +            AAA                      
                           AAA         +         AAA                          
                             AAAAA     +     AAAAA                            
  +---+--+---+---+--+---+---+---+-***********-+---+---+---+--+---+---+--+---+ 
-2                -1                                     1                 2

What type of user interface are you employing? There are (I think) some command-line versions of Maple that I have never used, so maybe it is that?

 

[LCKurtz]

What type of user interface are you employing? There are (I think) some command-line versions of Maple that I have never used, so maybe it is that?

I'm just using the Maple worksheet itself, not a command line version, the one I always use. I quit Maple and restarted it and the problem went away. It has shown up and disappeared on occasion in the past, for no apparent reason. I just Googled about it and apparently others have seen it, but I haven't seen any explanation yet.

 

[Raghav Gupta]

Try this in Wolfram Alpha.

http://m.wolframalpha.com/input/?i=graph of 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4)-2, from x=0 to x=7&x=4&y=10

That takes x from 0 to 7.

Thanks Sammy got it from graph.
Also thanks to Stephen, Ray and BvU. So x = 1,2,3,4 are asymptotic solutions.

 

[Stephen Tashi]

So x = 1,2,3,4 are asymptotic solutions.

To be clear, those values of x are the locations of vertical aymptotes. Those 4 values are NOT the 4 solutions to the equation.

 

[Raghav Gupta]

To be clear, those values of x are the locations of vertical aymptotes. Those 4 values are NOT the 4 solutions to the equation.

Yeah, I know.The question is solved