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Math Help - Double Sided Function Operations

  1. #1
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    Double Sided Function Operations

    Adding the same quantity to both sides of any equation is valid.
    So is subtracting, dividing and multiplying.

    My questions:
    - Is taking the radical of both sides a valid operation?
    - Is squaring/taking an arbitrary positive exponent of both sides a valid operation?
    - Is taking the reciprocal of both sides a valid operation?
    - Is taking the sin/cos/tan of both sides a valid operation?
    - Is taking the log of both sides to any base a valid operation?
    - Is taking the log of any number to the base of both sides a valid operation?
    - Is taking the limit as a variable approaches a certain value of both sides a valid operation? (<--THIS IS THE ONE I REALLY NEED HELP WITH)
    - Is differentiating (with respect to the same variable) both sides a valid operation?
    - Is integrating (with respect to the same variable) both sides a valid operation?
    - Is taking the absolute value of both sides a valid operation?
    - Is taking the arcsin/arccos/arctan of both sides a valid operation?

    Most importantly, can all the above be justified with proof or must they understood intuitively/axiomatically?

    Answer what you can my brothers. It is important I understand this. I am sorry if I posted in the wrong section.
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  2. #2
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    1.) Yes
    2.) Yes, however if there is a fractional exponent that you eliminate and that fractional exponent's denomator was even note that you will only get the positive part of that function. This is also not valid for inequalities.
    3.) Yes
    4.) Yes, unless it makes one undefined.
    5.) I don't know. I'm sorry. However I know that taking the common or natural log of both sides is legitamate as long as both sides are positive.
    6.) No, as one side may be negative.
    7.) I do not know. I am very sorry.
    8.) I've never seen a case where this is necesscary so I do not know.
    9.) View number 8.
    10.) I do not think so as it could eliminate negative signs on only one side.
    11.) Yes.

    They can all be justified by proofs that I do not know myself.
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  3. #3
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    What you're generally asking is this: are these functions (or operations) well-defined? That is, if you have a function f, is it always true that a = b implies f(a) = f(b), provided a and b are both in the domains? Another way of saying this is this: is it true that f is truly a function (single-valued)? If so, these sorts of operations you're talking about are all possible (subject to domain restrictions). The limit question is certainly true, because if a function has a limit, it is unique. That implies automatically that differentiation and integration are well-defined, because they're both limit processes.

    Make sense?
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  4. #4
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    Quote Originally Posted by Ackbeet View Post
    What you're generally asking is this: are these functions (or operations) well-defined? That is, if you have a function f, is it always true that a = b implies f(a) = f(b), provided a and b are both in the domains? Another way of saying this is this: is it true that f is truly a function (single-valued)? If so, these sorts of operations you're talking about are all possible (subject to domain restrictions). The limit question is certainly true, because if a function has a limit, it is unique. That implies automatically that differentiation and integration are well-defined, because they're both limit processes.

    Make sense?
    I understand that if f(x) is a function of x, and if a=b, then the statement f(a)=f(b) is not false. However, multiple solutions may arise out of the function f(a) and f(b).

    Example:
    Let f(x)=x^2

    Let x=y
    f(x)=f(y)
    x^2=y^2

    Solving for x gives x=y as well as x=-y (which was never a part of the premise).

    I understand generally what you mean but I need to be able to distinguish the function I have offered above from the functions that truly genuinely do not affect the truth value or the implications of a statement.
    Last edited by Skyrim; December 6th 2010 at 08:02 PM. Reason: fixing
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  5. #5
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    Solving for x gives x=y as well as x=-y (which was never a part of the premise).
    Right, but what are you doing here? You're inverting the square function. So if you let f(x)=x^{2}, and you assume that f(a)=f(b), then I agree that it might not be the case that a=b. However, that is not the meaning of the phrase "well-defined" as applied to functions. It's the difference in logic between the statements "If A then B" and "If B then A". One is the converse of the other, and, in general, neither implies the other. If it's true that A implies B, you cannot conclude that therefore, B implies A.

    So, getting back to a well-defined function and using our f(x)=x^{2} above, if it's true that a=b then it is always true that f(a)=f(b), or a^{2}=b^{2}. And that's true, because the square function is well-defined. A graphical way of thinking about a well-defined function is that it passes the vertical line test: any vertical line drawn on an xy plot of the function y=f(x) will touch the function in at most one place. So a circle is not a well-defined function. Technically, most people would say that it's not even a function.

    Thinking about things from an even higher level: whether or not you want to apply a particular function to an equation depends on a number of things. Do you need to be able to reverse all the steps in a series of computations? If so, then I would advise against applying functions that are not one-to-one (the graphical way of thinking about one-to-one functions is that they pass a horizontal line test, where any horizontal line intersects the graph in at most one place). The square function is not one-to-one, and hence applying the square function is not reversible. By "reversible", I mean that you can apply the function's inverse and exactly get to where you were before. When you apply functions to an equation that are not one-to-one, you can introduce spurious solutions to the equation. So you have to be careful when you square both sides of an equation, for example, that all the roots you find satisfy the original equation (before squaring).

    Does all this make sense? Any more questions?
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  6. #6
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    Quote Originally Posted by Ackbeet View Post
    Right, but what are you doing here? You're inverting the square function. So if you let f(x)=x^{2}, and you assume that f(a)=f(b), then I agree that it might not be the case that a=b. However, that is not the meaning of the phrase "well-defined" as applied to functions. It's the difference in logic between the statements "If A then B" and "If B then A". One is the converse of the other, and, in general, neither implies the other. If it's true that A implies B, you cannot conclude that therefore, B implies A.

    So, getting back to a well-defined function and using our f(x)=x^{2} above, if it's true that a=b then it is always true that f(a)=f(b), or a^{2}=b^{2}. And that's true, because the square function is well-defined. A graphical way of thinking about a well-defined function is that it passes the vertical line test: any vertical line drawn on an xy plot of the function y=f(x) will touch the function in at most one place. So a circle is not a well-defined function. Technically, most people would say that it's not even a function.

    Thinking about things from an even higher level: whether or not you want to apply a particular function to an equation depends on a number of things. Do you need to be able to reverse all the steps in a series of computations? If so, then I would advise against applying functions that are not one-to-one (the graphical way of thinking about one-to-one functions is that they pass a horizontal line test, where any horizontal line intersects the graph in at most one place). The square function is not one-to-one, and hence applying the square function is not reversible. By "reversible", I mean that you can apply the function's inverse and exactly get to where you were before. When you apply functions to an equation that are not one-to-one, you can introduce spurious solutions to the equation. So you have to be careful when you square both sides of an equation, for example, that all the roots you find satisfy the original equation (before squaring).

    Does all this make sense? Any more questions?
    Thanks, this does make sense. Now my question then is which functions are one-to-one and which are not and how can such things be proven? I know that cubic functions are generally not one-to-one unless their critical points and inflection points coincide. But I want to know if there is a way in general. If a=b, and if f(x)=x^3, then a^3 becomes b^3 and solving your only solution is a=b but what are some other possibilities in general?

    And sorry I can't write the math things the same way you can I don't have software or expertise.
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  7. #7
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    There are two equivalent ways to prove that a function is one-to-one.

    First method: Assume x, y are arbitrary members of the domain of f, and that f(x) = f(y). Prove that x = y.

    Second method: Assume x, y are arbitrary members of the domain of f, and that x\not=y. Prove that f(x)\not=f(y).

    These two methods are just logically the contrapositives of each other, which, unlike converses, are logically equivalent to each other.

    You can also convince yourself that a function is one-to-one (though this is NOT a proof), if it passes the horizontal line test mentioned above.

    One more way to show that a function is one-to-one is to show that its derivative is continuous everywhere on the domain, never changes sign, and is never zero except at isolated points. Equivalently, you can show one-to-one-ness if the derivative is continuous everywhere on the domain, and if x and y are two arbitrary members of the domain, then f'(x) f'(y)>0, except possibly at isolated points where the product might be zero.

    So there are a few methods of proving one-to-one-ness. Incidentally, the word "injective" is the same as "one-to-one". So a function is injective if-and-only-if it is one-to-one, by definition.

    Make sense?
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