# Tag Info

## Hot answers tagged calculus-and-analysis

8

Since Reduce doesn't seem to like the inequality, I tried FullSimplify with Assumptions instead. This works in three steps: differenceByTerm = SeriesCoefficient[(1 + x)^n - (1 + n x), {x, 0, m}] $$\cases{ 0 & m=0 \\ \binom{n}{m} & m>1 \\ 0 & \text{True} \\ }$$ FullSimplify[ differenceByTerm >= 0, Assumptions -> n ...

6

The antiderivative is correct, in the sense its derivative gives back the original integrand Clear[x] integrand = 4279/Sqrt[6817/10000 + 3183/10000*(1 + x)^3]; mmaResult = Integrate[integrand, x]; integrandBack = Simplify[D[mmaResult, x]]; Plot[{integrandBack, integrand}, {x, 0, 20}, PlotTheme -> "Detailed"] But the definite integral does not give ...

6

The expression in question is (we have replaced sigma by s) g = Integrate[ Exp[-(x - y)^2/(2 y s^2)] f[y] 1/Sqrt[2 \[Pi] y s^2], {y, 0, \[Infinity]}] $\int_0^{\infty } \frac{e^{-\frac{(x-y)^2}{2 s^2 y}} f[y]}{\sqrt{2 \pi } \sqrt{s^2 y}} \, dy$ First of all we notice that for y>0 "Limit[Exp[-(x - y)^2/(2 y s^2)] 1/Sqrt[2 \[Pi] y s^2], s -> 0] = ...

3

Here is another way: multi[f_, x_, a_, b_, n_] := Inactive@Integrate @@ {f @@ Array[x[#] &, {n}]}~Join~Array[{x[#], a[#], b[#]} &, {n}] I used Inactive because I don't know if the purpose is purely typesetting, or if you want it to try to evaluate (if so remove it, or use Activate later). multi[g, y, ymin, ymax, 10] (* Inactive[Integrate][ ...

3

Perhaps better: NMinimize[{m/(m (1 - c) + 1), 2^(m*(1 - c)) - 2.718/((m^.5)*2*3.14*c^(c*m + .5)*(1 - c)^((1 - c)*m + .5)) <= 0 && 0.006 <= c <= 1 && 4 <= m <= 2500}, {m, c}, MaxIterations -> 300] (* {1.11184, {m -> 4., c -> 0.350592}} *)

2

f[x_] := 3 x^4 + 8 x^3 - 24 x^2 - 48x + 19 s = 4; (*slope*) xs = x /. NSolve[f'[x] == s, x](*here are the x values where f'[x]=4*) lines = s (x - xs) + f[xs] (*here are the tangents*) Plot[{lines, f[x]}, {x, Min[xs], Max[xs]}]

2

Evaluating the indefinite integral works better. sol = Integrate[1/(w^4 + 2 (2 v^2 - 1) w^2 w1^2 + w1^4), w1, Assumptions -> {w > 0, 1 > v > 0}] (* (-(ArcTan[w1/(Sqrt[-1 + 2*v^2 + 2*v*Sqrt[-1 + v^2]]*w)]/ Sqrt[-1 + 2*v^2 + 2*v*Sqrt[-1 + v^2]]) - ArcTanh[w1/(Sqrt[1 - 2*v^2 + 2*v*Sqrt[-1 + v^2]]*w)]/ Sqrt[1 - 2*v^2 + 2*v*Sqrt[-1 + ...

2

You can split up the interval of integration: Integrate[DiracDelta[1 - x] DiracDelta[x] f[x], {x, 0, 1/2, 1}] (* 0 *) Still, it seems a bit inconvenient.

2

gradient[g_, vars_] := Table[D[g@@vars, vars[[j]]], {j, 1, Length[vars]}] system1[lstConst_, vars_] := Join[ Join@@ Table[gradient[lstConst[[j]], vars], {j, 1, Length[lstConst]}], Table[lstConst[[j]]@@vars,{j,1,Length[lstConst]}]]; system2[f_, lstConst_, vars_, lambda_] := Join[ gradient[f, vars] - Sum[ lambda[[j]]*gradient[lstConst[[j]], vars], ...

2

Your limit $$\lim _{n\rightarrow \infty }\dfrac {2^{n+1}-n-2}{2^{n}}=2$$ may be decomposed into the sum: $$\lim _{n\rightarrow \infty }\dfrac {2^{n+1}-n-2}{2^{n}}=\lim _{n\rightarrow \infty }\dfrac {2^{n+1}}{2^{n}}-\dfrac {n}{2^{n}}-\dfrac {2}{2^{n}}=2-0-0=2$$ And each limit: $$\lim _{n\rightarrow \infty }\dfrac {2^{n+1}}{2^{n}}=2$$ \lim _{n\rightarrow ...

2

Here's an inductive proof: Defining the function f[n_] := (1 + x)^n - (1 + n x) we have to prove that f[n] > 0 for n > 1 and x > -1. Now we observe that for f[n] we have the identity Simplify[f[n + 1] == (1 + x) f[n] + n x^2] (* True *) It is obvious "by eye" that f[n] > 0 because there are only positive quantities involved on the right ...

1

Using Sequence and Transpose to splice in the integration parameters: mi[f_, x_, a_, b_]:=Integrate[f @@ x, Sequence @@ Transpose@{x, a, b}]; Usage (exemplary): f[v_,w_,y_]:=v^2+w^2+y^2 x={x1,x2,x3}; a={a1,a2,a3}; b={b1,b2,b3}; mi[f,x,a,b] (* -(1/3)(a1-b1)(a2-b2)(a3-b3)(a1^2+a2^2+a3^2+a1 b1+b1^2+a2 b2+b2^2+a3 b3+b3^2) *)

1

If you use: x0 = (p^2 + k^2 + m^2 - (p0 - k0)^2)/(2 p k); f[x_] := (p^2 + k^2 - 2 p k x)/(x - x0); res=Integrate[f[x], {x, -1, 1}, PrincipalValue->True] you will presented with a result within a few minutes' time, however, there are lots of conditions spilled out to be fulfilled, which I left out here for clarity: -4 k p - (k0^2 - m^2 - 2 k0 p0 + ...

1

Indeed, the integral should give zero even with finite bounds. This workaround seems to give the desired result: Limit[ Integrate[ DiracDelta[1 - x] DiracDelta[x] f[x], {x, ϵ, 1}], ϵ -> 0] (* ==> 0 *) But it works only because it effectively cuts off the lower bound and thus the lower delta function.

1

With v10.0.2 it appears to work only for infinite bounds or using NIntegrate \$Version "10.0 for Mac OS X x86 (64-bit) (December 4, 2014)" Integrate[DiracDelta[1 - x] DiracDelta[x] f[x], {x, -Infinity, Infinity}] 0 Integrate[DiracDelta[x, 1 - x] f[x], {x, -Infinity, Infinity}] 0 NIntegrate[DiracDelta[1 - x] DiracDelta[x] f[x], {x, 0, ...

1

Your integral is too complicated for closed form. After fixing your syntax error, and integrating w.r.t. x only, you can see the result contains complex value and very complicated trig functions. Mathematica can't do the integration w.r.t. y at this stage f[x_] := 1/(a - c*x^2 + x*b I); g[y_] := 1/(a - c*y^2 + y*b I); h[z_] := 1/(a - c*z^2 + z*b I); j[x_, ...

1

Perhaps you will find a step-by-step solution helps you to understand how to approach a problem of this type. The function which defines the curve. f[x_] := 3 x^4 + 8 x^3 - 24 x^2 - 48 x + 19 The function giving the slope of the curve. slope[x_] = D[f[x], x] -48 - 48 x + 24 x^2 + 12 x^3 Finding the x values for which the slope is 4. xPts = x /. ...

1

It might be Errr... a bug or some singular points(not on the real axis) might be involved and Integrate chose a strange Integrate route. define f[x_] := Integrate[4279/Sqrt[0.6817 + 0.3183*(1 + x)^3], x] then fdat = Transpose@{Range[0, 20, 0.1], f[x] /. x -> Range[0, 20, 0.1]} plot the f generated by Integrate: ListLinePlot[{{#[[1]], ...

1

but why For the why part, you need little bit of math. You can always start by looking at series expansion of all terms expr = (Exp[x] - Exp[Sin[x]])/(x - Sin[x]); Limit[expr, x -> 0] (*1*) Now expand all in taylor (used 4 terms to make it easy to see) expr = (Normal@Series[Exp[x], {x, 0, 4}] - Normal@Series[Exp[Sin[x]], {x, 0, 4}])/(x - ...

1

First of all correcting the sign in the exponential we get f2 = Integrate[ Exp[r^2/(2 M^2)] (b^2 - 3 r^2)^2 (Abs[b^2 + r^2] - (b^2 - r^2))/( r (r^2 - b^2)), {r, 0, I \[Infinity]}, Assumptions -> {M > 0, b > 0] (* Out[278]= 18 b^2 M^2 + 4 b^4 E^(b^2/(2 M^2)) ExpIntegralEi[-(b^2/(2 M^2))] *) Taking the pricipal value has no influence as ...

1

Outline As you didn't provide boundary and initial conditions and the function pa'[t] this solution must be generic. Your equation for p[u,t] is linear (I guess pa'(t) means D[p[u,t],u]/.u->a) and can therefore be solved by standard mathematical methods once you provide the boundary and initial conditions. Physically it describes diffusion in a cylinder. A ...

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