New answers tagged calculus-and-analysis
7
Since your question's been answered, let me tell you about the handy listing of notations used by the Wolfram Functions site. In particular, if I scroll down to the "F" section of this page, you'll see an explanation that you are indeed looking at Kampé de Fériet's function.
11
It is Kampé de Fériet function, introduced in Joseph Kampé de Fériet, "La fonction hypergéométrique.", Mémorial des sciences mathématiques, Paris, Gauthier-Villars.
Its definition is given on Notations page:
and, in an alternative form, in Wikipedia:
$${}^{p+q}f_{r+s}\left(
\begin{matrix}
a_1,\cdots,a_p\colon b_1,b_1{}';\cdots;b_q,b_q{}'; \\
...
0
Another option is to use Replace within the Manipulate
Clear[f]
f = Function[x, x^k/(1 + x^k)];
Manipulate[Plot[f[x] /. k -> k1, {x, 0, 5}], {k1, 1, 10}]
5
Hopefully we're converging on the desired integral:
Assuming[(0 < x1 < x2 < 1),
Integrate[n (n - 1) (1 - y)^(n - 2), {y, 0, x2}, {x, 0, x1}]]
which has answer
(n x1 (-1 + (1 - x2)^n + x2))/(-1 + x2)
Though it may be that what you are after is:
Assuming[(0 < x1 < x2 < 1),
Integrate[n (n - 1) (1 - y)^(n - 2), {x, 0, x1}, {y, 0, ...
4
Let
f = (Sin[x^2] + Sin[y^2])/(x - y)
be the function in question.
As pointed out in the answers to this question, finding multivariable limits automatically computationally is full of pitfalls. The idea behind the function lim in this answer was to use Maximum and Minimum to find bounds on the function and apply the squeeze theorem. It fails here ...
1
Maybe another way to look at it is to change variables :
subs = {x -> Sqrt[a + b], y -> Sqrt[a - b]};
expr = FullSimplify[(Sin[x^2] + Sin[y^2])/(x - y) /. subs]
(* -((2 Cos[b] Sin[a])/(Sqrt[a - b] - Sqrt[a + b])) *)
This should make the path dependence of the limit a bit more explicit :
Limit[expr, a -> b]
(* (Sqrt[2] Cos[b] Sin[b])/Sqrt[b] *)
3
Well, Mathematica gives you the correct answer to your input. You first specify that x should be replaced by 0 and afterwards the limit of y->0 should be calculated. The correct answer to that is 0. And yes, this is exactly the same as taking the limit x->0 first and taking the limit y->0 afterwards.
If the double limit exists it is the same as taking the ...
2
If you enter it this way
Limit[(Sin[x]^2 + Sin[y]^2)/(x - y), {x -> y}]
(* Output: {DirectedInfinity[…]} *)
or alternatively you can try
Limit[(Sin[x]^2 + Sin[y]^2)/(x - y) /. x -> y, y -> 0]
(* Output: ComplexInfinity *)
In your input, you have used ReplaceAll, /., to first Replace all x's with zero, after which you find the limit as y -> 0. ...
6
Here's another way to proceed, using Derivative[], and sidestepping the use of a dummy variable:
LogDerivative[f_] := Derivative[1][Composition[Log, f]]
Test:
LogDerivative[Sin][x]
Cot[x]
LogDerivative[Gamma][x]
PolyGamma[0, x]
LogDerivative[#^3 &][x]
3/x
6
Your operator must depend on both function and variable - in analogy to D function:
logD[f_, x_] := D[f, x]/f
or an alternative definition:
logD[f_, x_] := D[Log[f], x]
Of course your variables of differentiation and in the function must agree. Test it:
logD[f[x], x]
Derivative[1][f][x]/f[x]
logD[Sin[x], x]
Cot[x]
f = x^2; logD[f, x]
...
8
Try this:
Integrate[(a + x)/((b^2 + (a + x)^2) (1 + c*(a - x)^2)), {x, -Infinity, Infinity},
Assumptions -> {a ∈ Reals, b ∈ Reals, c ∈ Reals, b != 0, c > 0}]
Which gives:
(-4 a b^2 c π + 2 a Sqrt[c] (1 + 4 a^2 c + b^2 c) π Abs[b])/
(Abs[b] + c (8 a^2 - 2 b^2 + (4 a^2 + b^2)^2 c) Abs[b])
Note that I've added additional assumptions so that ...
2
You can use :
Integrate[UnitStep[k - Sqrt[kx^2 + ky^2 + kz^2]],
{kx, -Infinity, Infinity}, {ky, -Infinity, Infinity}, {kz, -Infinity, Infinity},
Assumptions -> {k \[Element] Reals}]
(* -(4/3) k^3 \[Pi] (-1 + UnitStep[-k]) *)
Integrate[Boole[k - Sqrt[kx^2 + ky^2 + kz^2] >= 0],
{kx, -Infinity, Infinity}, {ky, -Infinity, Infinity}, {kz, -Infinity, ...
3
I'm not sure what you're looking for in a 'right' answer, but here are some ways to take (and view) the derivative of your function:
f'[x]
D[f[x], x]
Dt[f[x], x] // TraditionalForm
Dt[f[x], x] // FullSimplify // TraditionalForm
Check the documentation for D and Dt and pay close attention to syntax (ie. where all the punctuation and arguments go)
4
There is another approach that sometimes works better (gives closed-form expressions rather than recurrence relations):
In[1]:= InverseFourierTransform[(-I k)^n FourierTransform[1/(1 + x^2)^Log[2], x, k] , k, x]
Out[1]= (2^(-1 + n - 1/2 Log[1/x^2])
Abs[x]^-Log[2] ((-I)^
n ((1 + n) x Gamma[(1 + n)/2] Gamma[
n/2 + Log[2]] ...
5
There is another approach that sometimes works better (gives closed-form expressions rather than recurrence relations):
In[1]:= InverseFourierTransform[(-I k)^n FourierTransform[1/(1 + x^2)^Log[2], x, k] , k, x]
Out[1]= (2^(-1 + n - 1/2 Log[1/x^2])
Abs[x]^-Log[2] ((-I)^
n ((1 + n) x Gamma[(1 + n)/2] Gamma[
n/2 + Log[2]] ...
3
There is another approach that sometimes works better (gives closed-form expressions rather than recurrence relations):
In[1]:= InverseFourierTransform[(-I k)^n FourierTransform[1/(1 + x^2)^Log[2], x, k] , k, x]
Out[1]= (2^(-1 + n - 1/2 Log[1/x^2])
Abs[x]^-Log[2] ((-I)^
n ((1 + n) x Gamma[(1 + n)/2] Gamma[
n/2 + Log[2]] ...
2
f2[x_, mu2_, sigma2_] := 1/Sqrt[2*sigma2^2]*Exp[-Sqrt[2]*Abs[(x - mu2)/sigma2]]
Integrate[x*f2[x, mu, h], {x, -Infinity, Infinity},
Assumptions -> {Element[mu, Reals], h > 0}]
(*
mu
*)
Edit
Let's make what he did wrong crystal clear to the OP.
Should have used SetDelayed (:=) rather than Set (=) when defining f2.
Needed to have an ...
0
One way to circumvent it is like this
f2[x_, mu2_, sigma2_] :=
Piecewise[{{1/Sqrt[2*sigma2^2]*Exp[-Sqrt[2]*((x - mu2)/sigma2)], x - mu2 >= 0},
{1/Sqrt[2*sigma2^2]*Exp[-Sqrt[2]*((mu2 - x)/sigma2)], x - mu2 < 0}}]
Integrate[x*f2[x, mu, sigma], {x, -Infinity, Infinity}, Assumptions -> sigma > ...
8
Try this:
In[1]:= Integrate[BesselJ[0, x]/(x + BesselJZero[0, 1]), {x, -Infinity, Infinity}]
Out[1]= Pi StruveH[0, BesselJZero[0, 1]]
12
I know two approaches to this:
In[1]:= FullSimplify[SeriesCoefficient[ArcTan[y], {y, x, n}] n!, Element[n, Integers] && n > 0]
Out[1]= 1/2 I ((-I - x)^n - (I - x)^n) (1 + x^2)^-n Gamma[n]
and
In[2]:= FullSimplify[InverseFourierTransform[(-I k)^n FourierTransform[
ArcTan[x], x, k] , k, x], Element[n, Integers] && n > 0]
...
1
In[77]:= $Version
Out[77]= "9.0 for Microsoft Windows (64-bit) (January 25, 2013)"
In[73]:= Timing[All = Integrate[Abs[Sin[b + x]], {x, 0, 2*\[Pi]}]]
Out[73]= {81.073720, 4}
In[75]:= Timing[
FullSimplify@
Integrate[Abs@Sin[b + x], {x, 0, 2 Pi},
Assumptions -> {0 < b < 2 Pi}]]
(*4*)
Out[75]= {1.404009, 4}
1
FullSimplify@ Integrate[Abs@Sin[b + x] , {x, 0, 2 Pi}, Assumptions -> {0 < b < 2 Pi}]
(*
4
*)
5
As always there are several ways to improve the speed
Options Use MaxRecursions and MaxPoints
Method Try using different methods to obtain quickest solution.
Precalculate Use Block or Module to have some intermediate results only calculated once when required.
Analyze Very general advice: use debug features as AbsoluteTiming at several places to see, ...
5
The distance is $\sqrt{x^2+y^2}$, and the rate of change is the total derivative with respect to time which can be done as follows:
Clear[x, y, t];
y = x^2 + 1;
Simplify[Dt[Sqrt[x^2 + y^2]] /. Dt[x] -> 2]
(* ==> (6 x + 4 x^3)/Sqrt[1 + 3 x^2 + x^4] *)
Here I'm using the total derivative Dt instead of the simple D because Dt automatically assumes ...
0
I see no fundamental problem in using ND to answer all your questions. First I'll repeat the definition of your example function, then I do a single and a third partial derivative. Following that, I'll repeat the test of the accuracy for the exponential function:
g[x_?NumericQ, y_?NumericQ, z_?NumericQ] = x y z + x^2 y^2 z
(* ==> x y z + x^2 y^2 z *)
...
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