|
|
| Line 1: |
Line 1: |
| In [[calculus]], the [[indefinite integral]] of a given function (i.e., the set of all [[antiderivative]]s of the function) is only defined [[up to]] an additive constant, the '''constant of integration'''.<ref>{{cite book | last=Stewart | first=James | authorlink=James Stewart (mathematician) | title=Calculus: Early Transcendentals |publisher=[[Brooks/Cole]] | edition=6th | year=2008 | isbn=0-495-01166-5}}</ref><ref>{{cite book | last1=Larson | first1=Ron | authorlink=Ron Larson (mathematician)| last2=Edwards | first2=Bruce H. | title=Calculus | publisher=[[Brooks/Cole]] | edition=9th | year=2009 | isbn=0-547-16702-4}}</ref> This constant expresses an ambiguity inherent in the construction of antiderivatives. If a function <math>f(x)</math> is defined on an [[interval (mathematics)|interval]] and <math>F(x)</math> is an antiderivative of <math>f(x)</math>, then the set of ''all'' antiderivatives of <math>f(x)</math> is given by the functions <math>F(x) + C</math>, where ''C'' is an arbitrary constant. The constant of integration is sometimes omitted in [[lists of integrals]] for simplicity.
| | Getting an offshore account is one the best approaches for a individual to guard their money, plus one of the finest types of offshore accounts are called no identification offshore account. This sort of account doesn't require a person to show their individuality when opening the account, and in many instances it's completely fine for a individual to open the account under a name that is different than their own. There are a number of factors a person ought to know in regards to a zero identification offshore account such as the means of opening an account, and what such accounts are often employed for.<br><br>How-to start a zero ID account<br><br>The procedure of opening-up a zero ID account is excessively easy. In most cases a person can just enter a lender office, and feel the means of opening-up an account. This generally involves having an individual take note of the name of the account, after which selecting what alternatives they want for that account. An individual can then simply withdraw money from your account my writing down a secret code. Some banks that offer offshore accounts won't actually demand a person to visit the financial institution workplace.<br><br>What these accounts are used for<br><br>These types of accounts are usually utilized by individuals who wish to shield their finances from things such as lawsuits, or having them taken away by a government. This really is essential, because it is one of many methods an individual may ensure that their money simply doesn't get arrested for almost any purpose. Further Information [http://highlimitprepaidcard.blog.com/2014/08/26/high-limit-prepaid-card/ Get More Details]. |
| | |
| ==Origin of the constant==
| |
| | |
| The derivative of any constant function is zero. Once one has found one antiderivative <math>F(x)</math> for a function <math>f(x)</math>, adding or subtracting any constant ''C'' will give us another antiderivative, because <math>(F(x) + C)' = F\,'(x) + C\,' = F\,'(x)</math>. The constant is a way of expressing that every function has an infinite number of different antiderivatives.
| |
| | |
| For example, suppose one wants to find antiderivatives of <math>\cos(x)</math>. One such antiderivative is <math>\sin(x)</math>. Another one is <math>\sin(x)+1</math>. A third is <math>\sin(x)-\pi</math>. Each of these has derivative <math>\cos(x)</math>, so they are all antiderivatives of <math>\cos(x)</math>.
| |
| | |
| It turns out that adding and subtracting constants is the only flexibility we have in finding different antiderivatives of the same function. That is, all antiderivatives are the same up to a constant. To express this fact for cos(''x''), we write:
| |
| :<math>\int \cos(x)\,dx = \sin(x) + C.</math>
| |
| Replacing ''C'' by a number will produce an antiderivative. By writing ''C'' instead of a number, however, a compact description of all the possible antiderivatives of cos(''x'') is obtained. ''C'' is called the '''constant of integration'''. It is easily determined that all of these functions are indeed antiderivatives of <math>\cos(x)</math>:
| |
| :<math>\begin{align}
| |
| \frac{d}{dx}[\sin(x) + C] &= \frac{d}{dx}[\sin(x)] + \frac{d}{dx}[C] \\
| |
| &= \cos(x) + 0 \\
| |
| &= \cos(x)
| |
| \end{align}</math>
| |
| | |
| ==Necessity of the constant==
| |
| | |
| At first glance it may seem that the constant is unnecessary, since it can be set to zero. Furthermore, when evaluating [[definite integral]]s using the [[fundamental theorem of calculus]], the constant will always cancel with itself.
| |
| | |
| However, trying to set the constant equal to zero doesn't always make sense. For example, <math>2\sin(x)\cos(x)</math> can be integrated in at least three different ways:
| |
| | |
| :<math>\begin{align}
| |
| \int 2\sin(x)\cos(x)\,dx &=& \sin^2(x) + C &=& -\cos^2(x) + 1 + C &=& -\frac12\cos(2x) + C\\
| |
| \int 2\sin(x)\cos(x)\,dx &=& -\cos^2(x) + C &=& \sin^2(x) - 1 + C &=& -\frac12\cos(2x) + C\\
| |
| \int 2\sin(x)\cos(x)\,dx &=& -\frac12\cos(2x) + C &=& \sin^2(x) + C &=& -\cos^2(x) + C
| |
| \end{align}</math>
| |
| | |
| So setting ''C'' to zero can still leave a constant. This means that, for a given function, there is no "simplest antiderivative".
| |
| | |
| Another problem with setting ''C'' equal to zero is that sometimes we want to find an antiderivative that has a given value at a given point (as in an [[initial value problem]]). For example, to obtain the antiderivative of <math>\cos(x)</math> that has the value 100 at ''x'' = π, then only one value of ''C'' will work (in this case ''C'' = 100).
| |
| | |
| This restriction can be rephrased in the language of [[differential equations]]. Finding an indefinite integral of a function <math>f(x)</math> is the same as solving the differential equation <math>\frac{dy}{dx} = f(x)</math>. Any differential equation will have many solutions, and each constant represents the unique solution of a well-posed [[initial value problem]]. Imposing the condition that our antiderivative takes the value 100 at ''x'' = π is an initial condition. Each initial condition corresponds to one and only one value of ''C'', so without ''C'' it would be impossible to solve the problem.
| |
| | |
| There is another justification, coming from [[abstract algebra]]. The space of all (suitable) real-valued functions on the [[real number]]s is a [[vector space]], and the [[differential operator]] <math>\frac{d}{dx}</math> is a [[linear operator]]. The operator<math>\frac{d}{dx}</math> maps a function to zero if and only if that function is constant. Consequently, the [[kernel (algebra)|kernel]] of <math>\frac{d}{dx}</math> is the space of all constant functions. The process of indefinite integration amounts to finding a preimage of a given function. There is no canonical preimage for a given function, but the set of all such preimages forms a [[coset]]. Choosing a constant is the same as choosing an element of the coset. In this context, solving an [[initial value problem]] is interpreted as lying in the [[hyperplane]] given by the [[initial condition]]s.
| |
| | |
| ==Reason for a constant difference between antiderivatives ==
| |
| | |
| This result can be formally stated in this manner: Let <math>F:\mathbb{R}\rightarrow\mathbb{R}</math> and <math>G:\mathbb{R}\rightarrow\mathbb{R}</math> be two everywhere differentiable functions. Suppose that <math>F\,'(x) = G\,'(x)</math> for every real number ''x''. Then there exists a real number ''C'' such that <math>F(x) - G(x) = C</math> for every real number ''x''.
| |
| | |
| To prove this, notice that <math>[F(x) - G(x)]' = 0</math>. So ''F'' can be replaced by ''F''-''G'' and ''G'' by the constant function 0, making the goal to prove that an everywhere differentiable function whose derivative is always zero must be constant:
| |
| | |
| Choose a real number ''a'', and let <math>C = F(a)</math>. For any ''x'', the [[fundamental theorem of calculus]] says that
| |
| :<math>\begin{align}
| |
| \int_a^x 0\,dt &= F(x)-F(a)\\
| |
| &= F(x)-C,
| |
| \end{align}</math>
| |
| which implies that <math>F(x)=C</math>. So ''F'' is a constant function.
| |
| | |
| Two facts are crucial in this proof. First, the real line is [[connected space|connected]]. If the real line were not connected, we would not always be able to integrate from our fixed ''a'' to any given ''x''. For example, if we were to ask for functions defined on the union of intervals [0,1] and [2,3], and if ''a'' were 0, then it would not be possible to integrate from 0 to 3, because the function is not defined between 1 and 2. Here there will be ''two'' constants, one for each [[connected space|connected component]] of the [[function domain|domain]]. In general, by replacing constants with [[locally constant function]]s, we can extend this theorem to disconnected domains. For example, there are two constants of integration for <math>\textstyle\int dx/x</math> and infinitely many for <math>\textstyle\int \tan x\,dx,</math> so for example the general form for the integral of 1/''x'' is:<ref>"[http://golem.ph.utexas.edu/category/2012/03/reader_survey_logx_c.html Reader Survey: log|''x''| + ''C'']", Tom Leinster, ''The ''n''-category Café'', March 19, 2012</ref>
| |
| :<math>\int {1 \over x}\,dx = \begin{cases}\ln \left|x \right| + C^- & x < 0\\
| |
| \ln \left|x \right| + C^+ & x > 0
| |
| \end{cases}</math>
| |
| | |
| Second, ''F'' and ''G'' were assumed to be everywhere differentiable. If ''F'' and ''G'' are not differentiable at even one point, the theorem fails. As an example, let <math>F(x)</math> be the [[Heaviside step function]], which is zero for negative values of ''x'' and one for non-negative values of ''x'', and let <math>G(x)=0</math>. Then the derivative of ''F'' is zero where it is defined, and the derivative of ''G'' is always zero. Yet it's clear that ''F'' and ''G'' do not differ by a constant.
| |
| Even if it is assumed that ''F'' and ''G'' are everywhere continuous and [[almost everywhere]] differentiable the theorem still fails. As an example, take ''F'' to be the [[Cantor function]] and again let ''G'' = 0.
| |
| | |
| ==References==
| |
| {{reflist}}
| |
| | |
| [[Category:Integral calculus]]
| |
Getting an offshore account is one the best approaches for a individual to guard their money, plus one of the finest types of offshore accounts are called no identification offshore account. This sort of account doesn't require a person to show their individuality when opening the account, and in many instances it's completely fine for a individual to open the account under a name that is different than their own. There are a number of factors a person ought to know in regards to a zero identification offshore account such as the means of opening an account, and what such accounts are often employed for.
How-to start a zero ID account
The procedure of opening-up a zero ID account is excessively easy. In most cases a person can just enter a lender office, and feel the means of opening-up an account. This generally involves having an individual take note of the name of the account, after which selecting what alternatives they want for that account. An individual can then simply withdraw money from your account my writing down a secret code. Some banks that offer offshore accounts won't actually demand a person to visit the financial institution workplace.
What these accounts are used for
These types of accounts are usually utilized by individuals who wish to shield their finances from things such as lawsuits, or having them taken away by a government. This really is essential, because it is one of many methods an individual may ensure that their money simply doesn't get arrested for almost any purpose. Further Information Get More Details.