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

Attachments:

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 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 doesn’t make any difference whether t is a real, imaginary or complex number, the answer is always the same.

Thanks,

Mitch Sandlin

Attachments:

This is the code that works for that. Thank you!

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

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 didn’t 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

Attachments:

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.

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