Notation and Parenthesis Debate

[wvo77]

Hey guys, can you guys clear up the following for me.

• All of these notations mean the same thing: d/dx sin² (x), d/dx sin (x)² and d/dx (sin x)² ? And d/dx sin (x²) is something completely different.

• All of these notations mean the same thing : d/dx ln² (x), d/dx ln (x)² and d/dx (ln(x))² ? And d/dx ln(x²) is something completely different.

That being, said in a HW set the following occurred. In order to get d/dx of ln(5x^-3)^1/2 my prof. first transformed it by multiplying the exponents outside the parentheses with those inside and got: ln(5^1/2 x^-3/2). He says that with ln(5x^-3)^1/2 he means ln((5x^-3)^1/2). My issues is that if this is the case, then how come I can’t transform ln(x)² into ln(x²) by simply multiplying the exponent outside the ( ) with the exponent inside? I know that d/x ln(x)² does not equal d/x ln(x²), although my professor says they are!

Any help and insight would be great!


I know I could do it if I had ln(x)²

 

[pka]

Hey guys, can you guys clear up the following for me.

• All of these notations mean the same thing: d/dx sin² (x), d/dx sin (x)² and d/dx (sin x)² ? And d/dx sin (x²) is something completely different.

• All of these notations mean the same thing : d/dx ln² (x), d/dx ln (x)² and d/dx (ln(x))² ? And d/dx ln(x²) is something completely different.

Historically \(\displaystyle \sin^2(x)\) and \(\displaystyle [\sin(x)]^2\) mean the same thing.
The first was historically used up to the invention of computer algebra systems software c1970's.
The CAS generally use the second notation.

Now \(\displaystyle \sin^2(x)~\&~\sin(x^2)\) stand for two entirely different concepts. The first is the square of the sine of a number: \(\displaystyle {\sin ^2}\left( {\dfrac{\pi }{3}} \right) = \dfrac{3}{4}\) BUT \(\displaystyle \sin \left( {{{\left[ {\dfrac{\pi }{3}} \right]}^2}} \right) = \sin \left( {\dfrac{{{\pi ^2}}}{9}} \right) = 0.88967\)

In general function theory \(\displaystyle f^2(x)=[f(x)]^2\ne f(x^2)\).

 

[wvo77]

Historically \(\displaystyle \sin^2(x)\) and \(\displaystyle [\sin(x)]^2\) mean the same thing.
The first was historically used up to the invention of computer algebra systems software c1970's.
The CAS generally use the second notation.

Now \(\displaystyle \sin^2(x)~\&~\sin(x^2)\) stand for two entirely different concepts. The first is the square of the sine of a number: \(\displaystyle {\sin ^2}\left( {\dfrac{\pi }{3}} \right) = \dfrac{3}{4}\) BUT \(\displaystyle \sin \left( {{{\left[ {\dfrac{\pi }{3}} \right]}^2}} \right) = \sin \left( {\dfrac{{{\pi ^2}}}{9}} \right) = 0.88967\)

In general function theory \(\displaystyle f^2(x)=[f(x)]^2\ne f(x^2)\).

Thanks for information. That's what I thought.

But what about ln(x²) and ln(x)² ? I believe this is ambiguous..

 

[pka]

Thanks for information. That's what I thought.

But what about ln(x²) and ln(x)² ? I believe this is ambiguous..

I agree that it may appear to be ambiguous.
But I wound take \(\displaystyle \ln(x)^2=[\ln(x)]^2\) because most computer algebra systems would.

To see what I mean, look at this webpage.

 

[wvo77]

I agree that it may appear to be ambiguous.
But I wound take \(\displaystyle \ln(x)^2=[\ln(x)]^2\) because most computer algebra systems would.

To see what I mean, look at this webpage.

Exactly. Trust me, through Wolfram I confirmed my initially theory.

My issues is that to transform ln(5x^-3)^1/2 to ln(5^1/2 x^-3/2), like my prof did, you would be assuming that ln(5x^-3)^1/2 = ln((5x^-3)^1/2), while Wolfram reads it as [ln((5x^-3)^1/2)].

Any thoughts?

 

[lookagain]

Exactly. Trust me, through Wolfram I confirmed my initially theory.

My issues is that to transform ln(5x^-3)^1/2 to ln(5^1/2 x^-3/2), like my prof did, you would be assuming that ln(5x^-3)^1/2 = ln((5x^-3)^1/2), while Wolfram reads it as [ln((5x^-3)^1/2)].

Any thoughts?

"...you would be assuming that ln(5x^-3)^1/2 = ln((5x^-3)^1/2), * while Wolfram reads it as [ln((5x^-3)^1/2)]." **

* and ** are the same. The second one that has the brackets as part of it doesn't change anything about the value.

That being, said in a HW set the following occurred. In order to get d/dx of ln(5x^-3)^1/2my prof. first transformed it by multiplying the exponents outside the parentheses with those inside and got: ln(5^1/2 x^-3/2). He says that with ln(5x^-3)^1/2 he means ln((5x^-3)^1/2). My issues is that if this is the case, then how come I can’t transform ln(x)²into ln(x²) by simply multiplying the exponent outside the ( ) with the exponent inside?

Your professor needs to be careful and consistent. If he means the equivalent of \(\displaystyle ln((5x^{-3})^{1/2}), \ \) then he better write that, \(\displaystyle \ ln[(5x^{-3})^{1/2}], \ \ \)or something similar. \(\displaystyle \ \ \ \ \ \ \ \ \) \(\displaystyle ln^2(x) = [ln(x)]^2. \) \(\displaystyle \ \ \ \) Don't write \(\displaystyle "ln(x)^2."\) \(\displaystyle \ \ \ \) Either write \(\displaystyle "[ln(x)]^2"\) to show the square of the logarithm, or write \(\displaystyle "ln[(x)^2]" \ \ or \ \ "ln(x^2)" \ \) to represent the logarithm of the square of x.

 

[mmm4444bot]

But what about ln(x²) and ln(x)² ? I believe this is ambiguous.

I do not see any ambiguity because I interpret the symbolism ln(x) as function notation.

In other words, I see ln(x) as a single number (i.e., the output of the natural log function, when x is the input), so the square of this number (the function output) is ln(x)^2.

That's no different than squaring the output of function f or squaring the output of the cosine function, when the inputs are x: f(x)^2 and cos(x)^2.

The notations f(x^2) and cos(x^2) and ln(x^2) are clear, for those who understand function notation; the inputs are being squared because in function notation the input always appears between the parentheses.

Cheers :cool: