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Difference Between Inverse & Reciprocal

  1. Apr 22, 2008 #1
    Hello,
    can someone tell me the difference between reciprocal and the inverse. I have looked up the defenition and when I think I get a deep understanding I get confused. Thanks for all the help.

    Naicidrac
     
  2. jcsd
  3. Apr 22, 2008 #2

    symbolipoint

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    Maybe. What kind of inverse? Do you mean "multiplicative inverse" or do you mean "inverse function"?

     
  4. Apr 22, 2008 #3
    I think reciprocal is used for the scalar inverses, like 1/x. though, I am not neither native speaker nor a die-hard notation expert.
     
  5. Apr 22, 2008 #4

    HallsofIvy

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    "inverse" can apply to a number of different situations.

    In ordinary arithmetic the additive inverse is the negative: the additive inverse of 2 is -2.

    The multiplicative inverse is the reciprocal: the multiplicative inverse of 2 is [itex]\frac{1}{2}[/itex].

    If we are talking about functions, then the inverse function is the inverse with respect to "composition of functions": f(f-1(x))= x and f-1(f(x))= x.

    The word "inverse" can be applied to all of those (and more) but "reciprocal" only applies to multiplicative inverse.
     
  6. Apr 22, 2008 #5
    Ok awesome, Thanks for the clarity. I am thinking in terms of a trigonometric function like the difference between 1/sin(X) and sin^-1(X). What is the difference between these?
     
  7. Apr 22, 2008 #6

    ehj

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    1/sin(x) is the reciprocal of sin(x) so if sin(x) = 0,5 then 1/sin(x) = 2.
    sin^-1(x) is the inverse function of sin(x), sometimes also called arcsin(x).
     
  8. Apr 22, 2008 #7

    arildno

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    Out of sneaky, sleazy suggestiveness, mathematicians have adopted the convention [itex]f^{-1}(x)[/itex] for the inverse of a function, although if we think in terms of powers, this might be misunderstood as [itex](f(x))^{-1}=\frac{1}{f(x)}[/itex], i.e, the reciprocal of f, rather than the inverse function!

    Note in particular the nasty convention of the inverse trig functions, like [itex]\tan^{-1}(x)[/itex].

    Here, students are particularly prone to confuse the inverse trig with the reciprocal of the trig, because for all natural n's we choose to write [itex]\tan^{n}(x)[/itex], rather than the cumbersome [itex](\tan(x))^{n}[/tex]



    To make the issue even muddier, you may be unfortunate enough to see an iterated function like f(f(f(x))) written as [itex]f^{3}(x)[/itex] :smile:
     
    Last edited: Apr 22, 2008
  9. Feb 20, 2010 #8
    Reciprocal implies an equality. To reciprocate a smile means to smile back.
    While inverse implies an opposite. To invert a smile means to frown.
    However in mathematics it is often stated that the inverse of a number is the reciprocal of a number.
    I believe, however, this is incorrect usage.

    Very confusing when dealing with trig functions.
    First of all, can they truly be called functions?

    It seems tan^2 (x) is equivalent to tan(x) * tan(x).
    But again, you have to be careful, because this is different from tan(tan x) or even tan (x + x).

    These errors are similar to other distributing errors, such as,
    (3 x 3) x 2 = 6 x 6
    or (3 + 3)^2 = 9 + 9
    (which are false)
     
    Last edited: Feb 20, 2010
  10. Feb 20, 2010 #9
    Insidious!
     
    Last edited: Feb 20, 2010
  11. Feb 20, 2010 #10

    Mark44

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    Of course they can be called functions. Each one of the trig functions pairs a number in its domain with a number in its range. Did you mean the inverse trig functions when you said this? If so, the inverse of the function y = sin(x) is not itself a function, since each number in the domain of the inverse function maps to an infinite number of values in the range. To get around this problem, the domains of the trig functions are limited to an interval on which the trig function is one-to-one, thereby permitting an inverse that is a function.
    More precisely, they are identically equal by definition of the notation.
    I'm not aware of any special notation for collapsing tan(tan(x)). tan(x + x) = tan(2x), which would be difficult to confuse with tan2(x).
     
  12. Feb 20, 2010 #11
    To find the inverse of 2x = y, (we switch the x with the y)
    we use 2y = x OR y=x/2

    For the reciprocal of 2x = y (we switch the denominator with the numerator of y)
    that is 2x = 1/y OR y = 1/2x

    Remember a reciprocal of a number is any number you can multiply by that number to = 1.
    such as -1 and -1, or 12 and 1/12

    It may help to remember an inverse can be either - or + but the reciprocal will always be the same sign (thus it reciprocates the sign)

    Take the function, F(x) = 3x + 2
    y = 3x + 2 is changed to x = 3y+2 OR y = x/3 + 2
    The reciprocal of F(x) is 1/F(x).
    y = 3x+2 is changed to 1/y = 3x+2 OR y=1/(3x+2)

    If we set x to 0 we get -2 and 1/2, for the inverse and reciprocal functions respectively.
    However, if you solve for F(x) with you see that y = 2.
    The reciprocal of 2 is 1/2. The inverse is -2.


    This is what happens to exponents:
    When you invert x^N= y you get:
    y^N = x,
    y = x^(1/N)
    OR y= nth root of x

    When you reciprocate x^N=y you get:
    1/y = x^N,
    y = x^(-N)
    OR y=1/x^N

    Through inverting the exponent is reciprocated! This is very non-intuitive... or so you would think!

    tan(x) = sin (x) / cos (x)
    1/tan(x) = cos (x) / sin(x)
    tan(-1)(x) = ???
    So how do you find inverse trig functions?
    Well if you dedicate to memory your trig functions
    COS (0) = 1
    you know [COS (1)]^-1 = 0
    It is just reversing x and y of the function (cos (x) = y).

    if COS angle A is equal to X/R, then COS^(-1) of X/R is equal to angle A.
     
    Last edited: Feb 20, 2010
  13. Feb 20, 2010 #12
    Well, I have seen people try to factor/distribute with functions as if they were coefficients. I probably should have stated this.

    What I meant by asking if trigs are functions is: don't we just use a chart to find trig values? Without a formula could it be a true function?
     
  14. Feb 20, 2010 #13

    jgens

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    A formula isn't necessary since functions are defined in terms of collections of ordered pairs. More explicitly, a function [itex]f[/itex] is a collection of ordered pairs such that if [itex](a,b)[/itex] and [itex](a,c)[/itex] are both in [itex]f[/itex] then [itex]b = c[/itex]. Therefore, so long as that property is satisfied, then the trigonometric functions really are functions.

    And if you're really that caught up about formulas and equations, then how about this:

    [tex]\sin{(x)} = \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}[/tex]
     
  15. Feb 20, 2010 #14
    Ooo, pretty...
     
  16. Feb 22, 2010 #15
    Code (Text):
    but "reciprocal" only applies to multiplicative inverse.
    Well sort of but there is such a thing in engineering as the 'reciprocal theorem'. This is a totally different animal.

    I think the arcsin(x) v 1/sin(x) example sums it up very well too.
     
  17. Apr 23, 2011 #16
    Not to mention diabolical!

    I came here for primarily the same clarification. I'm just accepting it that in trig reciprocal is the same as a multiplicative reciprocal unless we're talking about identities then it's a reciprocal ID i.e. sin x = 1/csc x and inverses are i.e. sin¯ ¹ .

    JW
     
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