Tag Info

New answers tagged

1

Seems to be a bug. Maybe you can report it to Wolfram company. Here's just some observations with v8.0.4. If the assumptions are introduced by $Assumptions, the result is apparently incorrect: $Assumptions = R > 0 && f > 0 && m ∈ Integers; Integrate[Ebb[a] fb[a] + Ebc[a] fc[a], {a, 0, 2 π}] 0 but things will be different if we ...


6

The False setting can be useful when one wants an integral that is classically divergent. Or when one wants a result without provisos. A downside is greater chance of an incorrect result. A True setting is thus useful for the opposite, e.g. avoidance of finite results for divergent integrals. It can also be useful for more careful checking in multivariate ...


4

First, let us notice that the limit follows from the following: Lemma. $\displaystyle \lim_{u \rightarrow 0} {(1+u)^{1/4}-1 \over u} = {1 \over 4}$. For $${{(1+a\,x)^{1/4} - (1+b\,x)^{1/4}} \over {x}} = {{(1+a\,x)^{1/4} - 1} \over {x}} - {{(1+b\,x)^{1/4} - 1} \over {x}}$$ $$ = a\,{{(1+u)^{1/4} - 1} \over {u}} - b\,{{(1+v)^{1/4} - 1} \over {v}}\,,$$ ...


0

lst = Table[{a, NIntegrate[(-(a^2/2) + x (-x + Sqrt[a^2 + x^2]))* Tanh[\[Pi]*x], {x, 0, \[Infinity]}]}, {a, 0.2, 1, 0.05}]; ListPlot[lst]


3

This is at least how I might start such a problem: First define a function that calculates the numerical integral (using your definition) from 0 to some number: res[a_?NumericQ, xmax_?NumericQ] := res[a, xmax] = NIntegrate[ x Tanh[Pi x] Sqrt[x^2 + a^2] - (a^2/2 + x^2) Tanh[\[Pi] x], {x, 0, xmax}, WorkingPrecision -> 50] You might notice ...


10

One can simply use: Limit[((1 + a x)^(1/4) - (1 + b x)^(1/4))/x, x -> 0] to get the result. However Limit does know about the l'Hopital rule. Nonetheless there are different ways to go, e.g. let's use Series to write down the first few terms of the Taylor series of ((1 + a x)^(1/4) - (1 + b x)^(1/4))/x: Series[((1 + a x)^(1/4) - (1 + b x)^(1/4))/x, ...


5

This ordering is used in all functions in Mathematica, not just Integrate. Here's an example with Table: In[1]:= Table[f[i, j], {i, 3}, {j, 4}] Out[1]= {{f[1, 1], f[1, 2], f[1, 3], f[1, 4]}, {f[2, 1], f[2, 2], f[2, 3], f[2, 4]}, {f[3, 1], f[3, 2], f[3, 3], f[3, 4]}} The inner loop is according to j, the outer according i. In[2]:= ...


0

Here is a stepwise approach: sol = First@DSolve[f'[x] == -k (x - \[Mu]) f[x], f[x], x]; yields: {f[x] -> E^(k (-(x^2/2) + x \[Mu])) C[1]} In this case to get to the normal PDF: c1 = First@ Solve[Integrate[f[x] /. sol, {x, -Infinity, Infinity}, GenerateConditions -> False] == 1, C[1]] yields: {C[1] -> (E^(-((k \[Mu]^2)/2)) ...


1

Try this ... I changed your code a bit: eqn1 = f'[x] == -k (x - \[Mu]) f[x] eqn2 = 1/f \[DifferentialD]f == -k (x - \[Mu]) \[DifferentialD]x \[Integral]1/ f[x] \[DifferentialD]f[ x] == \[Integral]-k (x - \[Mu]) \[DifferentialD]x Exp[Log[f[x]]] == Exp[k (-(x^2/2) + x \[Mu])] (* you'll see it's the same as *) DSolve[eqn1, f[x], x] Gives for me: ...


0

To obtain the derivative of the integral, just put the derivative of the integrand inside: y[t_, x_] = t^2 x^2; ydt[t_, x_] = D[y[t, x], t]; g[t_?NumericQ] := NIntegrate[ydt[t, x], {x, 1, 2}] NIntegrate[g[t]*t, {t, 1, 2}]


5

I Let's write down an appropriate system we would like to solve, i.e. we are to maximize a^3 + b^3 + c^3 knowing that a + b + c == 5 and 1/a + 1/b + 1/c == 1/5, thus the most direct approach uses Maximize with adequate conditions: Maximize[{a^3 + b^3 + c^3, a + b + c == 5, 1/a + 1/b + 1/c == 1/5}, {a, b, c}] {125, {a -> 1, b -> 5, c -> -1}} ...


13

It's not a bug if you consider this behavior as a logical continuation of the following permissible syntax: D[a + b x^3, x, x, x] (* ==> 6 b *) D[a + b x^3, x, x] (* ==> 6 b x *) D[a + b x^3, x] (* ==> 3 b x^2 *) D[a + b x^3] (* ==> a + b x^3 *) The point is that a Sequence of variables is allowed following the first argument of D. And ...


0

Since the normal distribution is also built in to Mathematica, one could also do the calculation as follows: NProbability[ x - 1.96/Sqrt[5] < y < x + 1.96/Sqrt[5], {x \[Distributed] NormalDistribution[0, 1/Sqrt[5]], y \[Distributed] NormalDistribution[0, 1]} ] (* ==> 0.576386 *) As with NIntegrate versus Integrate, I use NProbability ...



Top 50 recent answers are included