New answers tagged calculus-and-analysis
1
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Analytical Solution in Generalized Heat Equation
just to add more to the comment, to help show why this is hard to solve analytically using separation of variables.
To solve using separation of variables we must be able to find the eigenvalues of ...
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1
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Intersection points of two-variable polynomials
Factoring shows they contain common factors.
...
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6
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4
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4
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Assume asymptotic value in a limit?
ClearAll[a, F, r, t, x]
Use TagSet to define UpValues for ...
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5
votes
Accepted
9
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Evaluate using Mathematica or otherwise the value of $f'(0)$
From definition of HypergeometricPFQ as Series representations we have:
$\underset{a\to 0}{\text{lim}}\frac{\partial }{\partial a}\, _3F_2\left(\frac{1}{2},\frac{1}{...
- 12.4k
2
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Evaluate using Mathematica or otherwise the value of $f'(0)$
Mathematica evaluates a closed form as a ConditionalExpression
Limit[(f[a] - f[0])/a, a -> 0]
which is probably not useful because it contains unevaluated ...
- 42.9k
3
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6
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4
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Accepted
Series expansion using binomial theorem in Mathematica
$Version
(* "13.2.1 for Mac OS X ARM (64-bit) (January 27, 2023)" *)
Clear["Global`*"]
f[x_] := (1 - a/x)^(1/3)
Do a series expansion about <...
- 139k
4
votes
Series expansion using binomial theorem in Mathematica
You are almost there. Try the following:
Series[(1 - (a/x))^(1/3), {a, 0, 2}] // Normal
(* 1 - a^2/(9 x^2) - a/(3 x) *)
Have fun!
- 37k
2
votes
I can't solve this problem
Sum[Log[((k + 1)*(k + 3))/(k + 2)^2], {k, 1, Infinity}]
yields -Log[3/2] or Log[2/3] or Log[2]-Log[3].
That is what Wolfram Mathematica and WolframAlpha do.
Of ...
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2
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2
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Accepted
Calculate wavelet coefficients with Mathematica
The a step function from -1/2 to 1/2 may be declared as:
rect[x_] = UnitStep[x + 1/2] - UnitStep[x - 1/2];
Plot[rect[x], {x, -1, 1}]
And your step function f:
<...
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1
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Evaluate $\int_{0}^{\pi/2} \cos^a (x)\sin(ax) dx$ using Mathematica
Workaround: using formula:
$$\sum _{j=1}^a \frac{\binom{a}{j} \sin (2 j x)}{2^a}=\cos ^a(x) \sin (a x)$$
We have:
$\int_0^{\frac{\pi }{2}} \cos ^a(x) \sin (a x) \, dx=2^{-1-a} \gamma +2^{-1-a} a \, ...
- 12.4k
3
votes
Green function and differential equation
Don't know why Mathematica GreenFunction doesn't evaluate.
Perhaps your last attempt might be elaborated a little bit further:
...
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1
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Evaluate $\int_{0}^{\pi/2} \cos^a (x)\sin(ax) dx$ using Mathematica
Nota Bene, for most functions Mathematica works in the complex domain unless explicitly told to work in another domain. That is done by stating the domain of the pertinent variables.
Since you believe ...
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5
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Evaluate $\int_{0}^{\pi/2} \cos^a (x)\sin(ax) dx$ using Mathematica
answer contains imaginary number i . How do we get a real answer?
Just because the answer contains $i$ does not necessarily mean the overall value is also complex. it depends on what is inside the ...
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2
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Evaluate $\int_{0}^{\pi/2} \cos^a (x)\sin(bx) dx$ using Mathematica
Clear["Global`*"]
When b == a the integral simplifies to
...
- 139k
4
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Evaluate $\int_{0}^{\pi/2} \cos^a (x)\sin(bx) dx$ using Mathematica
For $0\le x \le {\pi\over2}$, $a\ge0$, $b\ge0$,
we have $0 \le \cos x \le 1$ and $0 \le \sin bx \le bx$,
and therefore the integrand is bounded by
$0 \le \cos^a x \sin bx \le bx$ and the integral by $...
3
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Evaluate $\int_{0}^{\pi/2} \cos^a (x)\sin(bx) dx$ using Mathematica
This is getting too long to keep using comments. You need to use Limits for the special cases, then it works
...
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3
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Accepted
Is this a bug in IntegrateChangeVariables?
The wrong integral region is caused by failure of convergence of NMinValue.
...
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2
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Mathematica doesn't seem to be able to compute the Fourier transform of the Haar orthonormal basis over $L^2(\mathbb{R})$
Without loss of generality we may assume m==0 (see the translation property). Then in 13.2 on Windows 10 both
...
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1
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How to take the derivative of a function in `Im[]`?
We start from the relation $\Re[g'[x]]=(\Re[g[x]])'$ for real values of $x$.
Next, we apply the integral presentation of PolyLog (see Details in the documentation)
$...
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1
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Integrating a function over a surface integral
I've been try to perform calculations of exactly this type, and trying the different methods suggested. The energy surface I need is just as convoluted as the silicon Fermi surface shown in the OP and ...
1
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Could Mathematica analytically work with operators (such as $\nabla$) and vectors?
I think the expression should correctly be $(\nabla +P)\cdot( \nabla +P)$
Expression without brackets follows to $ \nabla \cdot \nabla +\nabla \cdot P+P \cdot \nabla + P \cdot P $
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0
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What is the equation to plot a vertical line?
Combined with Show, maybe you could use
ListLinePlot[{{2, -1}, {2, 1}}]
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2
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1
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How to print particular cases with Integrate?
yes but this integral exists even if u+v= +/- 1
Then do the following
...
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1
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Is there any hope to compute analytically $y=f(x)$ for which $g(x,y)=0$?
Here my numerical solution (inspired by @AlexeiBoulbitch useful comment):
upper contour
...
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0
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Doesn't analytically integrate sensibly let alone correctly
So to try to summarise everyone else's super problem solving and remove a distraction my erroneous intuition caused, the simple answer to my question is just that a real integration, at least a ...
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4
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Doesn't analytically integrate sensibly let alone correctly
x=0 evaluates indeterminate, for any positive x gives negative answer!
Lets fix the code first. You had {} inside the integrand and better use exact numbers. And ...
- 127k
2
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Accepted
Why earlier terms generated from AsymptoticDSolveValue change when increasing the order?
Here's what seems to be going on: variation of parameters is used to derive a particular solution:
...
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7
votes
Accepted
How to use Slater Type Orbitals as a basis functions in matrix method correctly?
Your orbitals are not orthogonal to each other: here I've normalized them with an explicit formula (using FindSequenceFunction to discover it):
...
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3
votes
Accepted
Mathematica calculates the parameter value when the first derivative is 0
I want to get the value of the corresponding parameter t when dil=0 or
dvc=0.
One way is to use the Solve command
...
- 127k
3
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Bug in integrate in in 13.1.0?
Another way of doing the integral.
g[x_] = 1/(x + 1)^2;
Integrate`InverseIntegrate[
1/(x + 1)^2 Exp[-t 1/(x + 1)^2], {x, 0, ∞}] /. t -> 3.
...
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6
votes
Accepted
Bug in integrate in in 13.1.0?
$Version
(* "13.1.0 for Mac OS X x86 (64-bit) (June 16, 2022)" *)
Clear["Global`*"]
g[x_] = 1/(x + 1)^2;
The incorrect result is
...
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0
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1
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Calculating an integral with a seemingly complicated integrand
At the first sight one can see that our integral depends on $u$ and $n$ and so we need not restrict them in advance but we should rather define our integral as a function of $u$ and $n$. Now we can ...
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1
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How to find the difference equation from system response or primitive equation?
There is an undocumented internal function that does this.
...
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4
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Why two the same integrals give different values?
Evaluating high-order polynomials is numerically unstable.
Example: exact evaluation followed by numericalization is stable,
Psi[7, 87] // N
(* -0.0271578 *)
...
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1
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Why earlier terms generated from AsymptoticDSolveValue change when increasing the order?
Inspired by the trick shown by Michael in the comment that when using using zero IC, spurious terms are gone, this is a function which will do this automatically. It only works with second order ode's ...
- 127k
3
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Why earlier terms generated from AsymptoticDSolveValue change when increasing the order?
$Version
(* "13.2.1 for Mac OS X x86 (64-bit) (January 27, 2023)" *)
ode = 2 x*y''[x] + (x + 1) y'[x] + 3 y[x] == x;
The exact solution with the ...
- 139k
5
votes
How to compute the residue of $f(z)=\frac{n/z}{z^{n}-1}$ to be $-n$ at $z=0$?
Regarding the comment: the residue at z=1 is clearly wrong
If a function, $f(z)$, has a pole of order $k$ at $z=z_0$ then
$$
\begin{equation}
\text{Res}(f,z_0)=\...
- 11.8k
0
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Speeding up the time of integration calculation
How about ParallelTable?
ParallelTable[NIntegrate[u2[K, P], {x, 0, a}, {y, 0, b}], {K, 1, n}, {P, 1, z}] // Timing
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2
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Accepted
Simple integral that becomes extremely complicated through a small modification
Leafcount might not be the best measure here. Why? because two expressions with very small difference can have the same leafcount. For example
...
- 127k
2
votes
Accepted
FunctionDomain and FunctionRange for x^2/x returns "True"
Observe that
g[x]
returns
x
So, when you write
FunctionDomain[g[x], x, Reals]
it is ...
- 11.8k
3
votes
Speeding up the time of integration calculation
Combining @UlrichNeumann's solution with analytic integration we get instantaneous results (0.6 seconds for 1000×1000 integrals) that remain accurate for very large $i$ and $j$:
...
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4
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Speeding up the time of integration calculation
The integral you are looking for might be separated into x-part and y-part.
Additionally introducing new variables \[Xi]=x/a, \[Eta]=y/b gives the following ...
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