Mixing
Area of a mushroom-shaped curve
Clash Royale CLAN TAG#URR8PPP
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Inspired by a discussion on this question, I discovered the following hybrid function:
The curves in red are defined as $$f(x)=expleft((sin x)^(sin x)^sin xright)$$ and the curves in blue are defined as $$g(x)=(sin x)^sin x$$
The result looks like the head of a mushroom (with a bit of decoration :)
Question: Consider just one 'mushroom head'. What is the area?
We can rewrite the problem as $$int_0^pileft[e^(sin x)^(sin x)^sin x-(sin x)^sin xright],dx$$ and we can see that it is symmetrical at $x=pi/2$, since $sinleft(fracpi2-xright)=sinleft(fracpi2+xright)$, so this is equivalent to $$2int_0^pi/2left[e^(sin x)^(sin x)^sin x-(sin x)^sin xright],dxtag1$$
Wolfram Alpha calculates this definite integral to be around $3.88407$ (not equal to, as pointed out in the comments.
So how should I tackle this integral? I do not anticipate a closed form, hence approximations would be fine.
Update: I have approximated the functions into simpler ones, to give a value of $3.86029$.
integration definite-integrals recreational-mathematics
asked Jul 17 at 19:54
TheSimpliFire
9,67361951
 |Â
show 3 more comments
up vote
8
down vote
favorite
Inspired by a discussion on this question, I discovered the following hybrid function:
The curves in red are defined as $$f(x)=expleft((sin x)^(sin x)^sin xright)$$ and the curves in blue are defined as $$g(x)=(sin x)^sin x$$
The result looks like the head of a mushroom (with a bit of decoration :)
Question: Consider just one 'mushroom head'. What is the area?
We can rewrite the problem as $$int_0^pileft[e^(sin x)^(sin x)^sin x-(sin x)^sin xright],dx$$ and we can see that it is symmetrical at $x=pi/2$, since $sinleft(fracpi2-xright)=sinleft(fracpi2+xright)$, so this is equivalent to $$2int_0^pi/2left[e^(sin x)^(sin x)^sin x-(sin x)^sin xright],dxtag1$$
Wolfram Alpha calculates this definite integral to be around $3.88407$ (not equal to, as pointed out in the comments.
So how should I tackle this integral? I do not anticipate a closed form, hence approximations would be fine.
Update: I have approximated the functions into simpler ones, to give a value of $3.86029$.
integration definite-integrals recreational-mathematics
asked Jul 17 at 19:54
TheSimpliFire
9,67361951
2
Is that $(sin x)^sin x$ or $sin(x^sin x)$ ?
– Dark Malthorp
Jul 17 at 19:56
4
@DarkMalthorp : Wouldn't $(sin x)^sin x$ be $sin^sin x x$?
– Eric Towers
Jul 17 at 19:57
2
Ah yes I suppose so. I personally completely hate that notation and so I never use it but thanks for the clarification
– Dark Malthorp
Jul 17 at 19:58
4
using the commandN[2*Int[Exp[Sin[x]^(Sin[x]^Sin[x]) ] - Sin[x]^(Sin[x]),x,0,Pi/2],40], WA returns3.8840669854123474566360415029651865604369...
– achille hui
Jul 17 at 20:04
2
+1 for the mushroom drawing
– David M.
Jul 17 at 20:13
 |Â
show 3 more comments
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Inspired by a discussion on this question, I discovered the following hybrid function:
The curves in red are defined as $$f(x)=expleft((sin x)^(sin x)^sin xright)$$ and the curves in blue are defined as $$g(x)=(sin x)^sin x$$
The result looks like the head of a mushroom (with a bit of decoration :)
Question: Consider just one 'mushroom head'. What is the area?
We can rewrite the problem as $$int_0^pileft[e^(sin x)^(sin x)^sin x-(sin x)^sin xright],dx$$ and we can see that it is symmetrical at $x=pi/2$, since $sinleft(fracpi2-xright)=sinleft(fracpi2+xright)$, so this is equivalent to $$2int_0^pi/2left[e^(sin x)^(sin x)^sin x-(sin x)^sin xright],dxtag1$$
Wolfram Alpha calculates this definite integral to be around $3.88407$ (not equal to, as pointed out in the comments.
So how should I tackle this integral? I do not anticipate a closed form, hence approximations would be fine.
Update: I have approximated the functions into simpler ones, to give a value of $3.86029$.
integration definite-integrals recreational-mathematics
asked Jul 17 at 19:54
TheSimpliFire
9,67361951
Inspired by a discussion on this question, I discovered the following hybrid function:
The curves in red are defined as $$f(x)=expleft((sin x)^(sin x)^sin xright)$$ and the curves in blue are defined as $$g(x)=(sin x)^sin x$$
The result looks like the head of a mushroom (with a bit of decoration :)
Question: Consider just one 'mushroom head'. What is the area?
We can rewrite the problem as $$int_0^pileft[e^(sin x)^(sin x)^sin x-(sin x)^sin xright],dx$$ and we can see that it is symmetrical at $x=pi/2$, since $sinleft(fracpi2-xright)=sinleft(fracpi2+xright)$, so this is equivalent to $$2int_0^pi/2left[e^(sin x)^(sin x)^sin x-(sin x)^sin xright],dxtag1$$
Wolfram Alpha calculates this definite integral to be around $3.88407$ (not equal to, as pointed out in the comments.
So how should I tackle this integral? I do not anticipate a closed form, hence approximations would be fine.
Update: I have approximated the functions into simpler ones, to give a value of $3.86029$.
integration definite-integrals recreational-mathematics
asked Jul 17 at 19:54
TheSimpliFire
9,67361951
edited Jul 22 at 16:07
asked Jul 17 at 19:54
TheSimpliFire
9,67361951
asked Jul 17 at 19:54
TheSimpliFire
9,67361951
asked Jul 17 at 19:54
TheSimpliFire
9,67361951
9,67361951
2
Is that $(sin x)^sin x$ or $sin(x^sin x)$ ?
– Dark Malthorp
Jul 17 at 19:56
4
@DarkMalthorp : Wouldn't $(sin x)^sin x$ be $sin^sin x x$?
– Eric Towers
Jul 17 at 19:57
2
Ah yes I suppose so. I personally completely hate that notation and so I never use it but thanks for the clarification
– Dark Malthorp
Jul 17 at 19:58
4
using the commandN[2*Int[Exp[Sin[x]^(Sin[x]^Sin[x]) ] - Sin[x]^(Sin[x]),x,0,Pi/2],40], WA returns3.8840669854123474566360415029651865604369...
– achille hui
Jul 17 at 20:04
2
+1 for the mushroom drawing
– David M.
Jul 17 at 20:13
 |Â
show 3 more comments
2
Is that $(sin x)^sin x$ or $sin(x^sin x)$ ?
– Dark Malthorp
Jul 17 at 19:56
4
@DarkMalthorp : Wouldn't $(sin x)^sin x$ be $sin^sin x x$?
– Eric Towers
Jul 17 at 19:57
2
Ah yes I suppose so. I personally completely hate that notation and so I never use it but thanks for the clarification
– Dark Malthorp
Jul 17 at 19:58
4
using the commandN[2*Int[Exp[Sin[x]^(Sin[x]^Sin[x]) ] - Sin[x]^(Sin[x]),x,0,Pi/2],40], WA returns3.8840669854123474566360415029651865604369...
– achille hui
Jul 17 at 20:04
2
+1 for the mushroom drawing
– David M.
Jul 17 at 20:13
2
2
Is that $(sin x)^sin x$ or $sin(x^sin x)$ ?
– Dark Malthorp
Jul 17 at 19:56
Is that $(sin x)^sin x$ or $sin(x^sin x)$ ?
– Dark Malthorp
Jul 17 at 19:56
4
4
@DarkMalthorp : Wouldn't $(sin x)^sin x$ be $sin^sin x x$?
– Eric Towers
Jul 17 at 19:57
@DarkMalthorp : Wouldn't $(sin x)^sin x$ be $sin^sin x x$?
– Eric Towers
Jul 17 at 19:57
2
2
Ah yes I suppose so. I personally completely hate that notation and so I never use it but thanks for the clarification
– Dark Malthorp
Jul 17 at 19:58
Ah yes I suppose so. I personally completely hate that notation and so I never use it but thanks for the clarification
– Dark Malthorp
Jul 17 at 19:58
4
4
using the command
N[2*Int[Exp[Sin[x]^(Sin[x]^Sin[x]) ] - Sin[x]^(Sin[x]),x,0,Pi/2],40], WA returns 3.8840669854123474566360415029651865604369...– achille hui
Jul 17 at 20:04
using the command
N[2*Int[Exp[Sin[x]^(Sin[x]^Sin[x]) ] - Sin[x]^(Sin[x]),x,0,Pi/2],40], WA returns 3.8840669854123474566360415029651865604369...– achille hui
Jul 17 at 20:04
2
2
+1 for the mushroom drawing
– David M.
Jul 17 at 20:13
+1 for the mushroom drawing
– David M.
Jul 17 at 20:13
 |Â
show 3 more comments
1 Answer
1
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oldest
votes
up vote
1
down vote
Curve approximation
Over the interval $[0,fracpi2]$, the function $a(x)=expleft(sin x^sin x^sin xright)$ can be approximated by the function $$alpha(x)=frac53sin x+1,$$ and similarly, the function $b(x)=sin x^sin x$ can be approximated by the function $$beta(x)=frac0.85xlnleft(0.66xright)e^x+1.$$ They are shown below, along with the original functions.
The function $alpha$ is easy to integrate. We get $$mathcal I_1=int_0^pi/2alpha(x),dx=left[x-frac53cos xright]_0^pi/2=fracpi2-frac53.$$
The function $beta$ is more difficult. Using WolframAlpha, we get $$mathcal I_2=int_0^pi/2beta(x),dxapprox1.30732.$$ To approximate this integral by hand, we could use the Taylor series for $ln$ and $exp$, but of course, this could only be limited to a number of terms (practically), as long rational functions are also extremely hard to integrate.
Hence the definite integral we want is $$int_0^pi a(x)-b(x),dxapprox2(mathcal I_1-mathcal I_2)=pi+frac103-2times1.30732=3.86029$$ with an error of around $0.61%$.
answered Jul 22 at 16:05
TheSimpliFire
9,67361951
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Curve approximation
Over the interval $[0,fracpi2]$, the function $a(x)=expleft(sin x^sin x^sin xright)$ can be approximated by the function $$alpha(x)=frac53sin x+1,$$ and similarly, the function $b(x)=sin x^sin x$ can be approximated by the function $$beta(x)=frac0.85xlnleft(0.66xright)e^x+1.$$ They are shown below, along with the original functions.
The function $alpha$ is easy to integrate. We get $$mathcal I_1=int_0^pi/2alpha(x),dx=left[x-frac53cos xright]_0^pi/2=fracpi2-frac53.$$
The function $beta$ is more difficult. Using WolframAlpha, we get $$mathcal I_2=int_0^pi/2beta(x),dxapprox1.30732.$$ To approximate this integral by hand, we could use the Taylor series for $ln$ and $exp$, but of course, this could only be limited to a number of terms (practically), as long rational functions are also extremely hard to integrate.
Hence the definite integral we want is $$int_0^pi a(x)-b(x),dxapprox2(mathcal I_1-mathcal I_2)=pi+frac103-2times1.30732=3.86029$$ with an error of around $0.61%$.
answered Jul 22 at 16:05
TheSimpliFire
9,67361951
add a comment |Â
up vote
1
down vote
Curve approximation
Over the interval $[0,fracpi2]$, the function $a(x)=expleft(sin x^sin x^sin xright)$ can be approximated by the function $$alpha(x)=frac53sin x+1,$$ and similarly, the function $b(x)=sin x^sin x$ can be approximated by the function $$beta(x)=frac0.85xlnleft(0.66xright)e^x+1.$$ They are shown below, along with the original functions.
The function $alpha$ is easy to integrate. We get $$mathcal I_1=int_0^pi/2alpha(x),dx=left[x-frac53cos xright]_0^pi/2=fracpi2-frac53.$$
The function $beta$ is more difficult. Using WolframAlpha, we get $$mathcal I_2=int_0^pi/2beta(x),dxapprox1.30732.$$ To approximate this integral by hand, we could use the Taylor series for $ln$ and $exp$, but of course, this could only be limited to a number of terms (practically), as long rational functions are also extremely hard to integrate.
Hence the definite integral we want is $$int_0^pi a(x)-b(x),dxapprox2(mathcal I_1-mathcal I_2)=pi+frac103-2times1.30732=3.86029$$ with an error of around $0.61%$.
answered Jul 22 at 16:05
TheSimpliFire
9,67361951
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Curve approximation
Over the interval $[0,fracpi2]$, the function $a(x)=expleft(sin x^sin x^sin xright)$ can be approximated by the function $$alpha(x)=frac53sin x+1,$$ and similarly, the function $b(x)=sin x^sin x$ can be approximated by the function $$beta(x)=frac0.85xlnleft(0.66xright)e^x+1.$$ They are shown below, along with the original functions.
The function $alpha$ is easy to integrate. We get $$mathcal I_1=int_0^pi/2alpha(x),dx=left[x-frac53cos xright]_0^pi/2=fracpi2-frac53.$$
The function $beta$ is more difficult. Using WolframAlpha, we get $$mathcal I_2=int_0^pi/2beta(x),dxapprox1.30732.$$ To approximate this integral by hand, we could use the Taylor series for $ln$ and $exp$, but of course, this could only be limited to a number of terms (practically), as long rational functions are also extremely hard to integrate.
Hence the definite integral we want is $$int_0^pi a(x)-b(x),dxapprox2(mathcal I_1-mathcal I_2)=pi+frac103-2times1.30732=3.86029$$ with an error of around $0.61%$.
answered Jul 22 at 16:05
TheSimpliFire
9,67361951
Curve approximation
Over the interval $[0,fracpi2]$, the function $a(x)=expleft(sin x^sin x^sin xright)$ can be approximated by the function $$alpha(x)=frac53sin x+1,$$ and similarly, the function $b(x)=sin x^sin x$ can be approximated by the function $$beta(x)=frac0.85xlnleft(0.66xright)e^x+1.$$ They are shown below, along with the original functions.
The function $alpha$ is easy to integrate. We get $$mathcal I_1=int_0^pi/2alpha(x),dx=left[x-frac53cos xright]_0^pi/2=fracpi2-frac53.$$
The function $beta$ is more difficult. Using WolframAlpha, we get $$mathcal I_2=int_0^pi/2beta(x),dxapprox1.30732.$$ To approximate this integral by hand, we could use the Taylor series for $ln$ and $exp$, but of course, this could only be limited to a number of terms (practically), as long rational functions are also extremely hard to integrate.
Hence the definite integral we want is $$int_0^pi a(x)-b(x),dxapprox2(mathcal I_1-mathcal I_2)=pi+frac103-2times1.30732=3.86029$$ with an error of around $0.61%$.
answered Jul 22 at 16:05
TheSimpliFire
9,67361951
answered Jul 22 at 16:05
TheSimpliFire
9,67361951
answered Jul 22 at 16:05
TheSimpliFire
9,67361951
answered Jul 22 at 16:05
TheSimpliFire
9,67361951
9,67361951
add a comment |Â
add a comment |Â
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2
Is that $(sin x)^sin x$ or $sin(x^sin x)$ ?
– Dark Malthorp
Jul 17 at 19:56
4
@DarkMalthorp : Wouldn't $(sin x)^sin x$ be $sin^sin x x$?
– Eric Towers
Jul 17 at 19:57
2
Ah yes I suppose so. I personally completely hate that notation and so I never use it but thanks for the clarification
– Dark Malthorp
Jul 17 at 19:58
4
using the command
N[2*Int[Exp[Sin[x]^(Sin[x]^Sin[x]) ] - Sin[x]^(Sin[x]),x,0,Pi/2],40], WA returns3.8840669854123474566360415029651865604369...– achille hui
Jul 17 at 20:04
2
+1 for the mushroom drawing
– David M.
Jul 17 at 20:13