Hi;
It doesn’t seem that the Norm[ ] command is applying the trig identity Sin^2 + Cos^2 = 1 to give the correct answer - please see attached.
Thanks,
Mitch Sandlin
What would you like to do:
Norm[2 Cos[t] - 2 Sin[t]]
I would have no idea what you would like it to do… It is not the conventional use of Norm…
Your notebook does not contain Sin[x]^2 + Cos[x]^2. Even if it did, as Sander says, it is not a convention that Norm simplifies trig expressions. However:
In[9]:= TrigReduce[Sin[x]^2 + Cos[x]^2]
Out[9]= 1
In[10]:= Simplify[Sin[x]^2 + Cos[x]^2]
Out[10]= 1
As I understand it, the Norm[ ] should be doing this - Sqrt[(-2 Sin[t])^2 + (2 Cos[t])^2 + (0)^2]. Basically, it should add the square the value for vectors x, y and z and then take the square root.
As you say, you need a vector, you gave it a formula, that simplify doesn’t work the way you want it to. Perhaps you want to normalize using an integral over a certain domain?
Ok, I also do not quite understand what the actual problem is. But it is relatively easy to force Norm to apply to a scalar, if we interpret the scalar as a “one-dimensional vector”, i.e. wrapping it in curly brackets. Using the original notebook posted above it looks like this:
f[t_] := 2 Cos[t] + 2 Sin[t] + 4;
FullSimplify[f'[t]/Norm[{f'[t]}]]
which gives
Note, that I needed to enclose f’[t] in “{}”; this is of course quite clumsy, because Abs[] would have achieved the same thing without the curly brackets. Obviously, this is not always one, as can be seen when we plot the function:
Plot[f'[t]/Norm[{f'[t]}], {t, 0, 2 Pi}]
Plotting the nominator and denominator separately, we see that the result makes sense:
Plot[{f'[t], Norm[{f'[t]}]}, {t, 0, 2 Pi}]
Cheers,
Marco
PS: As I said it would make more sense to use Abs here:
FullSimplify[f'[t]/Abs[f'[t]]]
gives
without the use of the artificial curly brackets…
Hi Sander,
It is my understanding that the Norm is a scalar length of a vector (or magnitude). It converts the vector into a scalar by summing the squares of the individual vector components and then takes the square root, and this is what I am assuming that the Norm[ ] command does in Mathematica.
My end goal is to normalize the vector, which I would do by dividing the individual vector values by the norm. Incidentally, I tried to use the Normalize[ ] command in Mathematica, and it didnt give me the expected results either see attached.
In the attached, the answers are somewhat correct. However, the denominators are not reduced. In this example, the denominators should simplify to 2.
Thanks so much, I certainly appreciate your time in helping me with this.
Mitch Sandlin
I have attached a copy of file where the Norm[ ] and Normalize[ ] commands give the expected answers. However, the attached file does not have any trig functions.
Hi;
Attached is yet another example of the Norm[ ] command not reducing to the correct answer when using Trig functions. As can be seen by inspection the answer should be SqRoot(29). However it appears that the subsequent Integrate[ ] command produced the correct answer. In using the Norm[ ] command, I have tried Simplify[ ], FullSimplify[ ] and TrigReduce[ ] commands and they all produce very similar answers to just using the Norm[ ] command by itself. I realize that I am using the Norm[ ] command in a very simplistic manner, but it still seems that it should produce the expected reduced answer.
Is there someone at Mathematica that can tell me what I am doing wrong, if I am in fact doing something wrong, or can this be reported as possibly broken code. By the way, I am using version 9 of Mathematica.
Thanks,
Mitch Sandlin
Dear Mitch,
I am sorry but I still do not understand why you are expecting this to evaluate to \sqrt{29} for all t. It is true that this should evaluate to \sqrt{29} if t was real, but if it wasn’t it is a mistake to make that particular simplification. For example:
FullSimplify[Norm[r'[t]] /. t -> 1]
gives \sqrt{29}, but if I substitute another perfectly valid number for t this expression does not evaluate to \sqrt{29}:
FullSimplify[Norm[r'[t]] /. t -> I]
So if, without further information, Mathematica were to reduce the equation to \sqrt{29}, that would be incorrect, because it does not hold for complex numbers in general. So I think that Mathematica does not simplify the equation in the way you wish it to do, because it would be incorrect in general. So I would think that this can definitely not be reported as broken code, but rather as expected and desired behaviour.
Or am I missing the point here?
Cheers,
Marco
As Marco points out you need to make the assumption that t is real, you can do so like this:
Simplify[... , t \[Element] Reals]
Hi Marco;
In your example, you are exactly correct.
However, my question concerns both the Norm[ ] and Normalize[ ] commands not applying the well know trig identity (Sin(t)^2 + (Cos(t))^2 = 1 in the final answer, even using the Simplify[ ], FullSimplify[ ] and TrigReduce[ ] commands. In my attached example (example3), I manually created the Norm[ ] command and the output can be found in output[15] which is very similar to output[14], which is the output from the Norm[ ] command. When I use the Simplify[ ] command with the manually created norm command I get the desired simplified answer in output (16). I cannot find any command that I can prefix to either the Norm[ ] and Normalize[ ] commands to get the trig identity to be applied to the final answer.
Incidentally, in the trig identity (Sin(t)^2 + (Cos(t))^2 = 1, it doesnt make any difference whether t is a real, imaginary or complex number, the answer is always the same.
Thanks,
Mitch Sandlin
Hi Sander;
I tried your last recommendation of using the Reals keyword, and all it did was remove the Abs in front of the Sin[ ] and Cos[ ]. It still did not apply the trig identity, which would have reduced the answer to SqRoot(29).
Thanks,
Mitch Sandlin
Hi Macro;
I tried the Simplify[ ] command with the Reals definition on both the Norm[ ] and Normalize[ ] commands without any change to either output - see attached. If I am doing something wrong, I certainly can’t figure it out.
Thanks,
Mitch Sandlin
Hi,
if I add
Simplify[Norm[normalize], t \[Element] Reals]
at the and of your Example3.nb notebook it gives 1 as expected.
Cheers,
Marco
Thanks so much, that worked great!
Mitch Sandlin
This is the code that works for that. Thank you!