Prove the given equation that involves indices

[chwala]

Homework Statement

Prove that;
$$(a-a^{-1})(a^{\frac {4}{3}} + a^{\frac{-2}{3}}) = \frac {a^2-a^{-2}}{a^{\frac {-1}{3}}}$$

Relevant Equations

understanding of indices

My approach;
$$(a-a^{-1})(a^{\frac {4}{3}} + a^{\frac {-2}{3}})
=(a-\frac {1}{a})(a^{\frac {4}{3}} +\frac {1}{a^{\frac {2}{3}}})
=(\frac{a^2-1}{a})(\frac{a^2+1}{a^{\frac {2}{3}}})
=\frac{a^4+a^2-a^2-1}{a^{\frac {5}{3}}}
=\frac{a^4-1}{a^{\frac {5}{3}}}$$
now at this point i multiplied both numerator and denominator by ##a^{-2}##
to realize;
$$\frac{a^4⋅a^{-2}-1⋅a^{-2}}{a^{\frac {5}{3}}⋅a^{-2}}=\frac {a^2-a^{-2}}{a^{\frac {-1}{3}}}$$

Now to my question, is there a different way of proving this without bringing in ##a^{-2}##?

 

[SammyS]

Homework Statement:: Prove that;
$$(a-a^{-1})(a^{\frac {4}{3}} + a^{\frac{-2}{3}}) = \frac {a^2-a^{-2}}{a^{\frac {-1}{3}}}$$
Relevant Equations:: understanding of indices

My approach;
$$(a-a^{-1})(a^{\frac {4}{3}} + a^{\frac {-2}{3}})
=(a-\frac {1}{a})(a^{\frac {4}{3}} +\frac {1}{a^{\frac {2}{3}}})
=(\frac{a^2-1}{a})(\frac{a^2+1}{a^{\frac {2}{3}}})
=\frac{a^4+a^2-a^2-1}{a^{\frac {5}{3}}}
=\frac{a^4-1}{a^{\frac {5}{3}}}$$
now at this point i multiplied both numerator and denominator by ##a^{-2}##
to realize;
$$\frac{a^4⋅a^{-2}-1⋅a^{-2}}{a^{\frac {5}{3}}⋅a^{-2}}=\frac {a^2-a^{-2}}{a^{\frac {-1}{3}}}$$

Now to my question, is there a different way of proving this without bringing in ##a^{-2}##?

Notice that ##\displaystyle \ \ \left(a^{(4/3)}+a^{(-2/3)}\right)= \left(a^{(\,(1/3)+1)}+a^{(\,(1/3)-1)}\right)##

##\displaystyle \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = a^{1/3}\left(a+a^{-1}\right)##

 

[Prof B]

An alternative, which is similar to SammyS's solution is to prove

Homework Statement:: Prove that;
$$(a-a^{-1})(a^{\frac {4}{3}} + a^{\frac{-2}{3}}) = \frac {a^2-a^{-2}}{a^{\frac {-1}{3}}}$$
Relevant Equations:: understanding of indices

An alternative, which is similar to SammyS's suggestion, is to prove

$$(a-a^{-1})(a^{\frac {4}{3}} + a^{\frac{-2}{3}})a^{\frac{1}{3}} = a^2-a^{-2}$$
Since a can't be 0, you can then divide through by $a^\frac{1}{3}$

By the way, superscripts are usually called exponents. Subscripts are indices.

 

[chwala]

An alternative, which is similar to SammyS's solution is to prove

$$(a-a^{-1})(a^{\frac {4}{3}} + a^{\frac{-2}{3}})a^{\frac{1}{3}} = a^2-a^{-2}$$
Since a can't be 0, you can then divide through by $a^\frac{1}{3}$

By the way, superscripts are usually called exponents. Subscripts are indices.

I am not getting your point here...kindly elaborate how the knowledge of ##a≠0## will help in the problem...and why are you talking of superscripts and subscripts?

 

[SammyS]

I am not getting your point here...kindly elaborate how the knowledge of ##a≠0## will help in the problem...and why are you talking of superscripts and subscripts?

I think it's because of the title of this thread versus what the actual topic is.

By the way, @Prof B is new to PF.

 

[Mark44]

you can then divide through by $a^\frac{1}{3}$

LaTeX here uses either two $ characters at each end for standalone equations/expressions, or two # characters at each end for inline equations/expressions.

 

[chwala]

I think it's because of the title of this thread versus what the actual topic is.

By the way, @Prof B is new to PF.

Maybe he needs some lessons on latex ...

 

[SammyS]

Maybe he needs some lessons on latex ...

I should have been clearer in that reply. I was referring to his final comment which was about subscripts, superscripts, exponents, and indices.

 

[chwala]

I should have been clearer in that reply. I was referring to his final comment which was about subscripts, superscripts, exponents, and indices.

He mentions that subscripts are indices... i am not getting that. Is that true? in my understanding i know that the term 'exponents, indices and power' can be used interchangeably in Mathematics to mean the same thing...it's true that a superscript may imply an exponent in a given math term ( e.g a variable or a number ).

 

[Steve4Physics]

He mentions that subscripts are indices... i am not getting that. Is that true? in my understanding i know that the term 'exponents, indices and power' can be used interchangeably in Mathematics to mean the same thing...it's true that a superscript may imply an exponent in a given math term ( e.g a variable or a number ).

Take x cubed (x³) as an example. At an introductory level you will often find the superscript ‘3’ referred to as a power, an exponent or an index. Informally, these terms are used interchangeably, though using ‘index’ is not strictly correct.

The correct use of ‘index’ (pl. indices) is to identity a particular value in (for example) a list). Think of the index at the back of a book – each entry is a ‘pointer’.

E.g. you measure the heights of 4 people. You could call the individual values h₁, h₂, h₃ and h₄. There is one index and the numbers (1, 2, 3 and 4) are values of the index - written as subscripts.

You may be interested to know that (confusingly) at a more advanced level (e.g. tensors), indices can by written as superscripts!

E.g. the contravariant components of a vector v could be written as v¹, v², v³ and v⁴ but the numbers are not powers. They are values of an index - simply labels like the 1, 2, 3 and 4 for heights.

If you wanted, say, to cube the 2nd component of the above vector, you would write (v²)³.

 

[chwala]

Take x cubed (x³) as an example. At an introductory level you will often find the superscript ‘3’ referred to as a power, an exponent or an index. Informally, these terms are used interchangeably, though using ‘index’ is not strictly correct.

The correct use of ‘index’ (pl. indices) is to identity a particular value in (for example) a list). Think of the index at the back of a book – each entry is a ‘pointer’.

E.g. you measure the heights of 4 people. You could call the individual values h₁, h₂, h₃ and h₄. There is one index and the numbers (1, 2, 3 and 4) are values of the index - written as subscripts.

You may be interested to know that (confusingly) at a more advanced level (e.g. tensors), indices can by written as superscripts!

E.g. the contravariant components of a vector v could be written as v¹, v², v³ and v⁴ but the numbers are not powers. They are values of an index - simply labels like the 1, 2, 3 and 4 for heights.

If you wanted, say, to cube the 2nd component of the above vector, you would write (v²)³.

Thanks @Steve4Physics ...but allow me to ask on why it was necessary in the context of the OP's question to define or rather re-define the terms subscript and superscripts i.e in post ##3##.

 

[Prof B]

LaTeX here uses either two $ characters at each end for standalone equations/expressions, or two # characters at each end for inline equations/expressions.

Thank you.

 

[Prof B]

I am not getting your point here...kindly elaborate how the knowledge of ##a≠0## will help in the problem

You can't divide by 0. You can divide by anything that isn't 0.

...and why are you talking of superscripts and subscripts?

Because exponents and indices are different things, so the words can't be used interchangeably. Steve4Physics gave a very good explanation.

 

[Mark44]

Because exponents and indices are different things, so the words can't be used interchangeably. Steve4Physics gave a very good explanation.

You didn't read Steve4Physics's post closely enough, in which he said that exponents and indexes/indices are often used synonymously.

(Emphasis added...)

At an introductory level you will often find the superscript ‘3’ referred to as a power, an exponent or an index.

Additionally, I have noticed that European members here often use "index" in the same way that I would use "exponent." Tomato, tomahto...

 

[chwala]

I am wondering why @Prof B keeps misreading posts...I have gone through his posts on other questions...I hope he's really not looking at physicsforums questions and phishing for solutions from other sites...then respond to same questions on the forum ...there's clearly a lack of keenness to details. Mythoughts...