The biggest value for sin(x)/x

[Stephanus]

Dear PF Forum,
Finally I have written a simple software to plot Sin(X)

It can also draw y = sin(x) + cos(x). But I don't know how to use this function in real life.

This software is very simple.
All you need to do is altering this short function

    case FDrawMode of
      0: Result:=Sin(X)+Cos(X);
      1: Result:=Sin(X);
      2: Result:=Cos(X);
      3: Result:=Sin(X)/X;
    end;

Don't forget to alter this line also.

  LS:=edtDrawMode.Items;
  LS.Clear;
  LS.Add('y = Sin(x)+Cos(x)');
  LS.Add('y = Sin(x)');
  LS.Add('y = Cos(x)');
  LS.Add('y = Sin(x)/x');

If you see the last line. LS.Add('y = Sin(x)/x');.
What if X is zero?
I cheat
Put catch exception in the code.

      try 
        Y:=ScaleY*Fc*                 CalcY(X+Sh); // case of sin(x)/x; if x = 0 then CalcY(X) will cause error.
        YY:=Height1-Round(Y);
        CV.Pixels[XX,YY]:=PlotColor;
      except
        // if error then don't draw anything
      end;

Funnily the computer still draw the curve as it is supposed to be.
And interesting enough. for sin(x)/x the biggest value is 1 for $\lim_{h \to 0} \frac{\sin(h)}{h}$.
But how to proof that the biggest value for sin(x)/(x) is lim h -> 0?
The graph shows that, but we just can't say. "Well the graph shows that, so the biggest value for sin(x)/(x) is 1",
Anyone interested to proof it?

 

[Stephanus]

I think the logical tought is this:
The biggest value for F(x)/x is if x is almost zero.
It makes sense, if x is 4 pi or 8 pi, the value will almost zero, too.
But if x is almost zero then Sin(x) is also almost zero, too.
Hmm, anyone can give a bettter opinon?

 

[micromass]

The function $\frac{sin(x)}{x}$ is not defined in $0$. Thus it never becomes $1$. The function has no biggest value.

 

[micromass]

Anyway, finding the biggest value (maximum) of a function is a simple exercise with derivatives. Any calculus book should cover this in their section of derivatives.

 

[Stephanus]

The function $\frac{sin(x)}{x}$ is not defined in $0$. Thus it never becomes $1$. The function has no biggest value.

So the "proof" that $lim_{h \to 0} \frac{\sin(h)}{h}$ is not a "proof"? Thanks, I don't want to argue. Just want to know what other thinks.

 

[Stephanus]

Anyway, finding the biggest value (maximum) of a function is a simple exercise with derivatives. Any calculus book should cover this in their section of derivatives.

"Exercise", I like that.

 

[micromass]

So the "proof" that $lim_{h \to 0} \frac{\sin(h)}{h}$ is not a "proof"? Thanks, I don't want to argue. Just want to know what other thinks.

I don't see how $\lim_{h\rightarrow 0} \frac{\sin(h)}{h}$ is at all relevant here.

 

[Stephanus]

I don't see how $\lim_{h\rightarrow 0} \frac{\sin(h)}{h}$ is at all relevant here.

Thanks micromass, I don't even see it at the slightest either. It's just an interesting graph, I wouldn't know the answer before the computer shows it. Just want confirmation.

 

[HallsofIvy]

You "just want confirmation" of what?

Your question is very confused. You ask about the "biggest value" of sin(x)/x.

That function does NOT HAVE a largest value. Its value, for any x, is less than 1 but comes arbitrarily close to 1.

Then you talk about the limit as x goes to 0 which is a completely different question! Yes, the limit, as x goes to 0, of sin(x)/x is 1. There are a number of different ways to show that depending on exactly which definition of sin(x) you are using. But that is NOT the value of the function at x= 0 because the function is NOT DEFINED at x= 0.
There is a "removable discontinuity" at x= 0. You could "remove" it by defining f(x)= sin(x)/x if x is not 0, f(0)= 1. But that would be a different function.

 

[jedishrfu]

Thanks micromass, I don't even see it at the slightest either. It's just an interesting graph, I wouldn't know the answer before the computer shows it. Just want confirmation.

Sometimes that is the case, in this case though you know the sin(x) and you know that its periodic however the x increases and reduces the amplitude of the sin(x) so you should expect it to dampen out and approach zero at very large x values. For me the interesting part is where x<1 and the combination of sin(x) / x complements one another to create the curve shown.

You should try other trig functions to see how they act like cos(x)/x or tan(x)/x or (sin(x)+cos(x))/x as well as other functions you learn about and develop a skill to be able to look at a function and sketch how it will look without the computer's help.

 

[Stephanus]

You "just want confirmation" of what?

Confirmation if I'm right or wrong. First I imagine what is the biggest value for sin(x)/x? I think it might, it might not be x = 0. But one thing for sure.
$\lim_{h \to 0} \frac{\sin(h)}{h}$ is 1
And x = pi/2; sin(x)/x is 0. So if we think linearly, it seems from x = 0 to pi/2 the value should decrease right? (or wrong?)
Then I intend to write a computer software, not to find this answer, but because of this:

https://www.physicsforums.com/threads/sine-cosine-limit-summary.827535/
Can you plot the sin and coz functions? Visually, the derivatives and limits are rather easy to see.

Tools? Only Microsoft Paint and Microsoft Excel. Of course I could make a small computer program to PutPixel(X,ScreenHeight-Sin(X/100*Pi)*100), but.. that will be tedious.

Then I have time this afternoon, so I wrote a simple program.
Part of the code.

    case FDrawMode of
      0: Result:=Sin(X)+Cos(X);
      1: Result:=Sin(X);
      2: Result:=Cos(X);
      3: Result:=Sin(X)/X; // ADDED LATER ON
    end;

Then you can see the result as the graph above about sin(x)/x.
"The computer DOES NOT show error if x=0??" Don't worry I cheat , see my post above.
Okay, so I see the graph, it seems that if x close to zero sin(x)/x is the highest.
But you know about computer variable. I use a very rough calculation, loop by 0.0001*pi step. And 0.0001 is very rough for math but not for computer variable.
So there remains some question.
1. Isn't there along the line from x = 0 to pi/2 a spike where sin(x)/x is bigger than 1?
2. Sin(x)/x looks (yes, "looks", because it's shown on the screen) highest at x = 0, but x can't be zero, right?
3. Is this true that the biggest value for sin(x)/x is $\lim_{x \to 0} \frac{\sin(x)}{x}$
But:

The function $\frac{sin(x)}{x}$ is not defined in $0$. Thus it never becomes $1$. The function has no biggest value.

Okay, so I'm satisfied enogh with micromass answer.

Your question is very confused. You ask about the "biggest value" of sin(x)/x.

That function does NOT HAVE a largest value. Its value, for any x, is less than 1 but comes arbitrarily close to 1.

Yes, micromass has answered it.

Then you talk about the limit as x goes to 0 which is a completely different question! Yes, the limit, as x goes to 0, of sin(x)/x is 1. There are a number of different ways to show that depending on exactly which definition of sin(x) you are using. But that is NOT the value of the function at x= 0 because the function is NOT DEFINED at x= 0.
There is a "removable discontinuity" at x= 0. You could "remove" it by defining f(x)= sin(x)/x if x is not 0, f(0)= 1. But that would be a different function.

Yes.

double SinXPerX1(double arc)
{
  return(Sin(arc)/arc);
}

double SinXPerX2(double arc)
{
  if(arc==0) return(1); else return(Sin(arc)/arc);
}

Agreed, it's different functions.

 

[micromass]

1. Isn't there along the line from x = 0 to pi/2 a spike where sin(x)/x is bigger than 1?

No.

2. Sin(x)/x looks (yes, "looks", because it's shown on the screen) highest at x = 0, but x can't be zero, right?

Right.

3. Is this true that the biggest value for sin(x)/x is $\lim_{x \to 0} \frac{\sin(x)}{x}$

No.

 

[Stephanus]

"1. Isn't there along the line from x = 0 to pi/2 a spike where sin(x)/x is bigger than 1?" No
"2. Sin(x)/x looks (yes, "looks", because it's shown on the screen) highest at x = 0, but x can't be zero, right?" Right
"3. Is this true that the biggest value for sin(x)/x is $\lim_{x \to 0} \frac{\sin(x)}{x}$"No

Come on micromass. I didn't ask that. I'm afraid I have already asked too many questions a day. But that WAS my doubt, and I'm already satisfied with this:

The function $\frac{sin(x)}{x}$ is not defined in $0$. Thus it never becomes $1$. The function has no biggest value.

But, thank you very much for your answer. It removes my doubts. I'm afraid I have asked too many questions.

 

[Stephanus]

Sometimes that is the case, in this case though you know the sin(x) and you know that its periodic however the x increases and reduces the amplitude of the sin(x) so you should expect it to dampen out and approach zero at very large x values. For me the interesting part is where x<1 and the combination of sin(x) / x complements one another to create the curve shown.

You should try other trig functions to see how they act like cos(x)/x or tan(x)/x or (sin(x)+cos(x))/x as well as other functions you learn about and develop a skill to be able to look at a function and sketch how it will look without the computer's help.

It seems Cos(x)/x similar to (Sin(x)+Cos(x))/x

 

[Stephanus]

You should try other trig functions to see how they act like cos(x)/x or tan(x)/x or (sin(x)+cos(x))/x as well as other functions you learn about and develop a skill to be able to look at a function and sketch how it will look without the computer's help.

Ahhh, I already

    case FDrawMode of
      0: Result:=Sin(X)+Cos(X);
      1: Result:=Sin(X);
      2: Result:=Cos(X);
      3: Result:=Sin(X)/X;
      4: Result:=Cos(X)/X; // added
      5: Result:=Sin(X)/Cos(X)/X; // has to change tan to sin/cos, no Tan function in Delphi
      6: Result:=(Sin(x)+Cos(x))/x
    end;

Alter it, compile it, run it, save the graph and upload it.
Hmm.., I didn't think that tan(x)/x will look like that. Picture it without computer or paper? Well, I'm not Descartes, but I'll try.

 

[Mark44]

IMO, the most important point here is the difference between a function's value (which might not exist) at a particular point and the limit of a function at that point. Two of your graphs are significant here: the graphs of $y = \frac{sin(x)}{x}$ and $y = \frac{tan(x)}{x}$.

Both graphs appear to show a point at (0, 1). In fact, neither function is defined when x = 0. The graphing software (the program you wrote in Delphi) plots a number of points and then connects the dots.

Even though both functions are undefined at x = 0, both functions have limits that exist as x approaches 0. In mathematical notation $\lim_{x \to 0}\frac{sin(x)}{x} = 1$ and $\lim_{x \to 0}\frac{tan(x)}{x} = 1$. It's very easy to show that the latter limit can be gotten once you know the first limit.
$\lim_{x \to 0}\frac{tan(x)}{x} = \lim_{x \to 0}\frac{sin(x)}{cos(x) \cdot x} = \lim_{x \to 0}\frac{1}{cos(x)} \cdot \lim_{x \to 0}\frac{sin(x)}{x}$.

The first limit is clearly 1, since cos(0) = 1, and we know that the second limit is 1.

 

[Stephanus]

Hi Mark44, it's been a while since you answer my post in: https://www.physicsforums.com/threads/proof-limit-derivatave-of-sin-x.826864/page-3#post-5198643

IMO, the most important point here is the difference between a function's value (which might not exist) at a particular point and the limit of a function at that point.

Thanks a lot! A good point! Function value which might not exist vs the limit of that function in that point.

Two of your graphs are significant here: the graphs of $y = \frac{sin(x)}{x}$ and $y = \frac{tan(x)}{x}$.
Both graphs appear to show a point at (0, 1). In fact, neither function is defined when x = 0. The graphing software (the program you wrote in Delphi) plots a number of points and then connects the dots.

Yes both are undefined at x = 0.
Just plot the dots not connect it.
I put something like it in the software.

try // begin catch error
  Y = sin(x)/X; // calculate Y
  PuPixel(X,Y); // Draw the result
except
  // catch any error
  // don't do anything
end;

The original code is at my first post above.

Even though both functions are undefined at x = 0, both functions have limits that exist as x approaches 0.

Yes, yes I know. You've given me this lecture since 2 weeks ago. Thanks

In mathematical notation $\lim_{x \to 0}\frac{sin(x)}{x} = 1$ and $\lim_{x \to 0}\frac{tan(x)}{x} = 1$. It's very easy to show that the latter limit can be gotten once you know the first limit.

Proof why tan(x)/x = 1? Perhaps later. For now, I'm already satisfied that I can proof why $\lim_{h \to 0} \frac{\sin(h)}{h} = 1$

$\lim_{x \to 0}\frac{tan(x)}{x} = \lim_{x \to 0}\frac{sin(x)}{cos(x) \cdot x} = \lim_{x \to 0}\frac{1}{cos(x)} \cdot \lim_{x \to 0}\frac{sin(x)}{x}$.
The first limit is clearly 1, since cos(0) = 1, and we know that the second limit is 1.

Ahh, that's the proof? Very simple actually.