Hot answers tagged calculus-and-analysis
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]
...
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]
...
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, ...
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]] ...
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 ...
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]] ...
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 ...
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 ...
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)
2
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 ...
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. ...
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, ...
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] *)
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}
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