# The biggest value for sin(x)/x

Tags:
1. Aug 15, 2015

### 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
Code (Text):
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.
Code (Text):
LS:=edtDrawMode.Items;
LS.Clear;

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.
Code (Text):
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?

2. Aug 15, 2015

### 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?

3. Aug 15, 2015

### micromass

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

4. Aug 15, 2015

### micromass

Staff Emeritus
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.

5. Aug 15, 2015

### Stephanus

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.

6. Aug 15, 2015

### Stephanus

"Exercise", I like that.

7. Aug 15, 2015

### micromass

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

8. Aug 15, 2015

### Stephanus

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.

9. Aug 15, 2015

### HallsofIvy

Staff Emeritus
You "just want confirmation" of what?

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.

10. Aug 15, 2015

### Staff: Mentor

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.

11. Aug 15, 2015

### Stephanus

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:
Then I have time this afternoon, so I wrote a simple program.
Part of the code.
Code (Text):
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:
Okay, so I'm satisfied enogh with micromass answer.

Yes.
Code (C):
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.

12. Aug 15, 2015

### micromass

Staff Emeritus
No.

Right.

No.

13. Aug 15, 2015

### Stephanus

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

14. Aug 15, 2015

### Stephanus

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

15. Aug 15, 2015

### Stephanus

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

16. Aug 15, 2015

### Staff: Mentor

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.

17. Aug 15, 2015

### 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
Thanks a lot! A good point! Function value which might not exist vs the limit of that function in that point.
Yes both are undefined at x = 0.
Just plot the dots not connect it.
I put something like it in the software.
Code (Text):
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.
Yes, yes I know. You've given me this lecture since 2 weeks ago. Thanks
Proof why tan(x)/x = 1? Perhaps later. For now, I'm already satisfied that I can proof why $\lim_{h \to 0} \frac{\sin(h)}{h} = 1$
Ahh, that's the proof? Very simple actually.