Constant factor rule in differentiation; constant

Thursday, May 29th, 2008

In calculus, the constant factor rule in differentiation allows you to take constants outside a derivative and concentrate on differentiating the function of x itself. This is a part of the linearity of differentiation.

Suppose you have a function

<math>g(x) = k \cdot f(x).</math>

where k is a constant.

Use the formula for differentiation from first principles to obtain:

<math>g’(x) = \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}</math>
<math>g’(x) = \lim_{h \to 0} \frac{k \cdot f(x+h) - k \cdot f(x)}{h}</math>
<math>g’(x) = \lim_{h \to 0} \frac{k(f(x+h) - f(x))}{h}</math>
<math>g’(x) = k \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \mbox{(*)}</math>
<math>g’(x) = k \cdot f’(x).</math>

This is the statement of the constant factor rule in differentiation, in Lagrange’s notation for differentiation.

In Leibniz’s notation, this reads

<math>\frac{d(k \cdot f(x))}{dx} = k \cdot \frac{d(f(x))}{dx}.</math>

If we put k=-1 in the constant factor rule for differentiation, we have:

<math>\frac{d(-y)}{dx} = -\frac{dy}{dx}.</math>


Comment on proof

Note that for this statement to be true, k must be a constant, or else the k can’t be taken outside the limit in the line marked (*).

If k depends on x, there is no reason to think k(x+h) = k(x). In that case the more complicated proof of the product rule applies.

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Derivative of a constant; function for

Thursday, December 20th, 2007

In calculus, the derivative of a constant function is zero. (A constant function is one that does not depend on the independent variable, such as f(x) = 7.)

The rule can be justified in various ways. The derivative is the slope of the tangent to the given function’s graph, and the graph of a constant function is a horizontal line, whose slope is zero. Alternatively, one can use the limit definition of the derivative. The difference quotient
<math>\frac{f(x+h)-f(x)}{h}</math> is zero for every h, and so therefore must be the limit of this quotient as h tends to zero, that is, f’(x).


Antiderivative of zero

A partial converse to this statement is the following:

If a function has a derivative of zero on an interval, it must be constant on that interval.

This is not a consequence of the original statement, but follows from the Mean value theorem. It can be generalized to the statement that

If two functions have the same derivative on an interval, they must differ by a constant,

or

If g is an antiderivative of f on and interval, then all antiderivatives of f on that interval are of the form g(x)+C, where C is a constant.

From this follows a weak version of the second Fundamental theorem of calculus: if f is continuous on [a,b] and f = g’ for some function g, then

<math>\int_a^b f(x) dx = g(b) - g(a)</math>.

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De Bruijn-Newman constant; constant the

Saturday, September 29th, 2007

The De Bruijn-Newman constant, denoted by Λ, is a mathematical constant and is defined via the zeros of a certain function H(λ, z), where λ is a real parameter and z is a complex variable. H has only real zeros if and only if λ ≥ Λ. The constant is closely connected with Riemann’s hypothesis on the zeros of the general Euler-Riemann’s ζ-function. In brief, the Riemann hypothesis is equivalent to the conjecture that Λ ≤ 0.

De Bruijn showed in 1950 that H has only real zeros if λ ≥ 1/2,
and moreover, that if H has only real zeros for some λ, H also has only real zeros if λ is replaced by any larger value. Newman proved in 1976 the existence of a constant Λ for which the “if and only if” claim holds; and this then implies that Λ is unique. Newman conjectured that Λ ≥ 0, an intriguing counterpart to the Riemann hypothesis. Serious calculations on lower bounds for Λ have been made since 1988 and—as can be seen from the table—are still being made:

Year Lower bound on Λ
1988 −50
1991 −5
1990 −0.385
1994 −4.379 · 10−6
1993 −5.895 · 10−9
2000 −2.7 · 10−9

SInce <math> H(\lambda , z) </math> is just the Fourier transform of
<math> F(e^{\lambda x}\Phi) </math> then H has the Wiener-Hopf representation:

<math> \xi (1/2+iz)= A\sqrt \pi (\lambda)^{-1} \int_{-\infty}^{\infty}dx e^{\frac{-1}{4\lambda}(x-z)^{2}} H(\lambda , x) </math>

which is only valid for lambda positive or 0, it can be seen that in the limit lambda tends to zero then <math> H(0,x)=\xi(1/2+ix) </math> for the case Lambda is negative then H is defined so:

<math> H(z,\lambda)=B\sqrt \pi (\lambda)^{-1} \int_{-\infty}^{\infty}dx e^{\frac{-1}{4\lambda}(x-z)^{2}} \xi(1/2+ix) </math>

Where A and B are real constant.


External links

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