Proving the definition of $arcsin x$ using analysis
My goal is to prove that the function $arcsin:left [ -1,1 right ]rightarrow mathbb{R}$ can be defined as
$$x mapsto arcsin x equiv int_{0}^{x}frac{1}{sqrt{1-t^2}} dt,$$
which is odd and continuous. Assume nothing about the sine function is known.
Showing the function is continuous on $(-1,1)$ just follows from the definition.
Showing the function is odd should be as simple as showing that
$$f(-x)=-f(x).$$
Plugging this in gives
$$f(-x)=int_{0}^{-x}frac{1}{sqrt{1-t^2}} dt=-int_{-x}^{0}frac{1}{sqrt{1-t^2}} dt.$$
I am not too sure what to do next.
Lastly, I think the bulk of the question is to prove that the function we are integrating is Riemann integrable. That is, it is a valid definition. I am having trouble on this part as well because I am not sure how to construct my partition $P$ such that
$$U(f;P)-L(f;P)<epsilon.$$
I know this a lot- I am pretty confused! Can anyone provide me some feedback and help?
Thanks so much in advance!
real-analysis integration trigonometry
asked Nov 30 at 5:38
MathIsLife12
563111
add a comment |
My goal is to prove that the function $arcsin:left [ -1,1 right ]rightarrow mathbb{R}$ can be defined as
$$x mapsto arcsin x equiv int_{0}^{x}frac{1}{sqrt{1-t^2}} dt,$$
which is odd and continuous. Assume nothing about the sine function is known.
Showing the function is continuous on $(-1,1)$ just follows from the definition.
Showing the function is odd should be as simple as showing that
$$f(-x)=-f(x).$$
Plugging this in gives
$$f(-x)=int_{0}^{-x}frac{1}{sqrt{1-t^2}} dt=-int_{-x}^{0}frac{1}{sqrt{1-t^2}} dt.$$
I am not too sure what to do next.
Lastly, I think the bulk of the question is to prove that the function we are integrating is Riemann integrable. That is, it is a valid definition. I am having trouble on this part as well because I am not sure how to construct my partition $P$ such that
$$U(f;P)-L(f;P)<epsilon.$$
I know this a lot- I am pretty confused! Can anyone provide me some feedback and help?
Thanks so much in advance!
real-analysis integration trigonometry
asked Nov 30 at 5:38
MathIsLife12
563111
2
continuous functions are Riemann integrable.
– Lord Shark the Unknown
Nov 30 at 5:40
How would one construct the epsilon-delta continuity proof for that function, though?
– MathIsLife12
Nov 30 at 5:43
Are you just trying to prove that $f(x) = int_0^x frac{1}{sqrt{1-t^2}}text dt$ is a continuous odd function on the interval $[-1,1]$?
– AlexanderJ93
Nov 30 at 5:49
1
@AlexanderJ93 Yes. The question is worded weirdly but that is what it comes down to. I am a little stuck.
– MathIsLife12
Nov 30 at 5:51
You are on the right track, just let $t to -t$ in the last integral and simplify.
– Mustafa Said
Nov 30 at 6:02
add a comment |
My goal is to prove that the function $arcsin:left [ -1,1 right ]rightarrow mathbb{R}$ can be defined as
$$x mapsto arcsin x equiv int_{0}^{x}frac{1}{sqrt{1-t^2}} dt,$$
which is odd and continuous. Assume nothing about the sine function is known.
Showing the function is continuous on $(-1,1)$ just follows from the definition.
Showing the function is odd should be as simple as showing that
$$f(-x)=-f(x).$$
Plugging this in gives
$$f(-x)=int_{0}^{-x}frac{1}{sqrt{1-t^2}} dt=-int_{-x}^{0}frac{1}{sqrt{1-t^2}} dt.$$
I am not too sure what to do next.
Lastly, I think the bulk of the question is to prove that the function we are integrating is Riemann integrable. That is, it is a valid definition. I am having trouble on this part as well because I am not sure how to construct my partition $P$ such that
$$U(f;P)-L(f;P)<epsilon.$$
I know this a lot- I am pretty confused! Can anyone provide me some feedback and help?
Thanks so much in advance!
real-analysis integration trigonometry
asked Nov 30 at 5:38
MathIsLife12
563111
My goal is to prove that the function $arcsin:left [ -1,1 right ]rightarrow mathbb{R}$ can be defined as
$$x mapsto arcsin x equiv int_{0}^{x}frac{1}{sqrt{1-t^2}} dt,$$
which is odd and continuous. Assume nothing about the sine function is known.
Showing the function is continuous on $(-1,1)$ just follows from the definition.
Showing the function is odd should be as simple as showing that
$$f(-x)=-f(x).$$
Plugging this in gives
$$f(-x)=int_{0}^{-x}frac{1}{sqrt{1-t^2}} dt=-int_{-x}^{0}frac{1}{sqrt{1-t^2}} dt.$$
I am not too sure what to do next.
Lastly, I think the bulk of the question is to prove that the function we are integrating is Riemann integrable. That is, it is a valid definition. I am having trouble on this part as well because I am not sure how to construct my partition $P$ such that
$$U(f;P)-L(f;P)<epsilon.$$
I know this a lot- I am pretty confused! Can anyone provide me some feedback and help?
Thanks so much in advance!
real-analysis integration trigonometry
real-analysis integration trigonometry
asked Nov 30 at 5:38
MathIsLife12
563111
asked Nov 30 at 5:38
MathIsLife12
563111
edited Nov 30 at 5:40
asked Nov 30 at 5:38
MathIsLife12
563111
asked Nov 30 at 5:38
MathIsLife12
563111
asked Nov 30 at 5:38
MathIsLife12
563111
563111
2
continuous functions are Riemann integrable.
– Lord Shark the Unknown
Nov 30 at 5:40
How would one construct the epsilon-delta continuity proof for that function, though?
– MathIsLife12
Nov 30 at 5:43
Are you just trying to prove that $f(x) = int_0^x frac{1}{sqrt{1-t^2}}text dt$ is a continuous odd function on the interval $[-1,1]$?
– AlexanderJ93
Nov 30 at 5:49
1
@AlexanderJ93 Yes. The question is worded weirdly but that is what it comes down to. I am a little stuck.
– MathIsLife12
Nov 30 at 5:51
You are on the right track, just let $t to -t$ in the last integral and simplify.
– Mustafa Said
Nov 30 at 6:02
add a comment |
2
continuous functions are Riemann integrable.
– Lord Shark the Unknown
Nov 30 at 5:40
How would one construct the epsilon-delta continuity proof for that function, though?
– MathIsLife12
Nov 30 at 5:43
Are you just trying to prove that $f(x) = int_0^x frac{1}{sqrt{1-t^2}}text dt$ is a continuous odd function on the interval $[-1,1]$?
– AlexanderJ93
Nov 30 at 5:49
1
@AlexanderJ93 Yes. The question is worded weirdly but that is what it comes down to. I am a little stuck.
– MathIsLife12
Nov 30 at 5:51
You are on the right track, just let $t to -t$ in the last integral and simplify.
– Mustafa Said
Nov 30 at 6:02
2
2
continuous functions are Riemann integrable.
– Lord Shark the Unknown
Nov 30 at 5:40
continuous functions are Riemann integrable.
– Lord Shark the Unknown
Nov 30 at 5:40
How would one construct the epsilon-delta continuity proof for that function, though?
– MathIsLife12
Nov 30 at 5:43
How would one construct the epsilon-delta continuity proof for that function, though?
– MathIsLife12
Nov 30 at 5:43
Are you just trying to prove that $f(x) = int_0^x frac{1}{sqrt{1-t^2}}text dt$ is a continuous odd function on the interval $[-1,1]$?
– AlexanderJ93
Nov 30 at 5:49
Are you just trying to prove that $f(x) = int_0^x frac{1}{sqrt{1-t^2}}text dt$ is a continuous odd function on the interval $[-1,1]$?
– AlexanderJ93
Nov 30 at 5:49
1
1
@AlexanderJ93 Yes. The question is worded weirdly but that is what it comes down to. I am a little stuck.
– MathIsLife12
Nov 30 at 5:51
@AlexanderJ93 Yes. The question is worded weirdly but that is what it comes down to. I am a little stuck.
– MathIsLife12
Nov 30 at 5:51
You are on the right track, just let $t to -t$ in the last integral and simplify.
– Mustafa Said
Nov 30 at 6:02
You are on the right track, just let $t to -t$ in the last integral and simplify.
– Mustafa Said
Nov 30 at 6:02
add a comment |
1 Answer
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Powers and sums are continuous, and compositions of continuous functions are continuous, so the integrand is continuous on $(-1,1)$ hence RI, by the FTC, $sin^{-1}(x)$ is continuous. For oddness, use the change of variables $u=-x$.
Assume $xge 1/2$. Because the integrand is positive, the integral increases in $x$, so the change of variables $u=1-t^2$ expresses value of $|arcsin 1-arcsin x| $ as
$$ int_x^1 frac {text{d}u}{2sqrt u sqrt{1-u}}le frac 1{sqrt 2}int_{x}^1frac {text{d}u}{sqrt{1-u}}=sqrt 2sqrt{1-x}to 0 quad text{as $xto 1$}$$
Thus, $arcsin x$ is continuous at $1$. The same procedure can be applied at $-1$.
answered Nov 30 at 6:05
Guacho Perez
3,88911131
This is good for the open interval $(-1,1)$. However, you need to show the endpoints separately. Consider $g(x) = frac{1}{1-|x|}$ continuous on $(-1,1)$ but $int_0^x g(t)text dt$ not defined for $x in {-1,1}$.
– AlexanderJ93
Nov 30 at 6:29
1
@AlexanderJ93 Edited my post.
– Guacho Perez
Nov 30 at 7:00
Another approach to deal with endpoints of $[-1,1]$ is to note the equality $$int_{0}^{x}sqrt{1-t^2},dt=frac{xsqrt{1-x^2}}{2}+frac{1}{2}int_{0}^{x}frac{dt}{sqrt{1-t^2}}$$ valid for $xin(-1,1)$. Taking limits as $xto pm 1$ shows that the integral in question converges if $x=pm 1$.
– Paramanand Singh
Nov 30 at 17:19
add a comment |
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Powers and sums are continuous, and compositions of continuous functions are continuous, so the integrand is continuous on $(-1,1)$ hence RI, by the FTC, $sin^{-1}(x)$ is continuous. For oddness, use the change of variables $u=-x$.
Assume $xge 1/2$. Because the integrand is positive, the integral increases in $x$, so the change of variables $u=1-t^2$ expresses value of $|arcsin 1-arcsin x| $ as
$$ int_x^1 frac {text{d}u}{2sqrt u sqrt{1-u}}le frac 1{sqrt 2}int_{x}^1frac {text{d}u}{sqrt{1-u}}=sqrt 2sqrt{1-x}to 0 quad text{as $xto 1$}$$
Thus, $arcsin x$ is continuous at $1$. The same procedure can be applied at $-1$.
answered Nov 30 at 6:05
Guacho Perez
3,88911131
This is good for the open interval $(-1,1)$. However, you need to show the endpoints separately. Consider $g(x) = frac{1}{1-|x|}$ continuous on $(-1,1)$ but $int_0^x g(t)text dt$ not defined for $x in {-1,1}$.
– AlexanderJ93
Nov 30 at 6:29
1
@AlexanderJ93 Edited my post.
– Guacho Perez
Nov 30 at 7:00
Another approach to deal with endpoints of $[-1,1]$ is to note the equality $$int_{0}^{x}sqrt{1-t^2},dt=frac{xsqrt{1-x^2}}{2}+frac{1}{2}int_{0}^{x}frac{dt}{sqrt{1-t^2}}$$ valid for $xin(-1,1)$. Taking limits as $xto pm 1$ shows that the integral in question converges if $x=pm 1$.
– Paramanand Singh
Nov 30 at 17:19
add a comment |
Powers and sums are continuous, and compositions of continuous functions are continuous, so the integrand is continuous on $(-1,1)$ hence RI, by the FTC, $sin^{-1}(x)$ is continuous. For oddness, use the change of variables $u=-x$.
Assume $xge 1/2$. Because the integrand is positive, the integral increases in $x$, so the change of variables $u=1-t^2$ expresses value of $|arcsin 1-arcsin x| $ as
$$ int_x^1 frac {text{d}u}{2sqrt u sqrt{1-u}}le frac 1{sqrt 2}int_{x}^1frac {text{d}u}{sqrt{1-u}}=sqrt 2sqrt{1-x}to 0 quad text{as $xto 1$}$$
Thus, $arcsin x$ is continuous at $1$. The same procedure can be applied at $-1$.
answered Nov 30 at 6:05
Guacho Perez
3,88911131
This is good for the open interval $(-1,1)$. However, you need to show the endpoints separately. Consider $g(x) = frac{1}{1-|x|}$ continuous on $(-1,1)$ but $int_0^x g(t)text dt$ not defined for $x in {-1,1}$.
– AlexanderJ93
Nov 30 at 6:29
1
@AlexanderJ93 Edited my post.
– Guacho Perez
Nov 30 at 7:00
Another approach to deal with endpoints of $[-1,1]$ is to note the equality $$int_{0}^{x}sqrt{1-t^2},dt=frac{xsqrt{1-x^2}}{2}+frac{1}{2}int_{0}^{x}frac{dt}{sqrt{1-t^2}}$$ valid for $xin(-1,1)$. Taking limits as $xto pm 1$ shows that the integral in question converges if $x=pm 1$.
– Paramanand Singh
Nov 30 at 17:19
add a comment |
Powers and sums are continuous, and compositions of continuous functions are continuous, so the integrand is continuous on $(-1,1)$ hence RI, by the FTC, $sin^{-1}(x)$ is continuous. For oddness, use the change of variables $u=-x$.
Assume $xge 1/2$. Because the integrand is positive, the integral increases in $x$, so the change of variables $u=1-t^2$ expresses value of $|arcsin 1-arcsin x| $ as
$$ int_x^1 frac {text{d}u}{2sqrt u sqrt{1-u}}le frac 1{sqrt 2}int_{x}^1frac {text{d}u}{sqrt{1-u}}=sqrt 2sqrt{1-x}to 0 quad text{as $xto 1$}$$
Thus, $arcsin x$ is continuous at $1$. The same procedure can be applied at $-1$.
answered Nov 30 at 6:05
Guacho Perez
3,88911131
Powers and sums are continuous, and compositions of continuous functions are continuous, so the integrand is continuous on $(-1,1)$ hence RI, by the FTC, $sin^{-1}(x)$ is continuous. For oddness, use the change of variables $u=-x$.
Assume $xge 1/2$. Because the integrand is positive, the integral increases in $x$, so the change of variables $u=1-t^2$ expresses value of $|arcsin 1-arcsin x| $ as
$$ int_x^1 frac {text{d}u}{2sqrt u sqrt{1-u}}le frac 1{sqrt 2}int_{x}^1frac {text{d}u}{sqrt{1-u}}=sqrt 2sqrt{1-x}to 0 quad text{as $xto 1$}$$
Thus, $arcsin x$ is continuous at $1$. The same procedure can be applied at $-1$.
answered Nov 30 at 6:05
Guacho Perez
3,88911131
edited Nov 30 at 7:00
answered Nov 30 at 6:05
Guacho Perez
3,88911131
answered Nov 30 at 6:05
Guacho Perez
3,88911131
answered Nov 30 at 6:05
Guacho Perez
3,88911131
3,88911131
This is good for the open interval $(-1,1)$. However, you need to show the endpoints separately. Consider $g(x) = frac{1}{1-|x|}$ continuous on $(-1,1)$ but $int_0^x g(t)text dt$ not defined for $x in {-1,1}$.
– AlexanderJ93
Nov 30 at 6:29
1
@AlexanderJ93 Edited my post.
– Guacho Perez
Nov 30 at 7:00
Another approach to deal with endpoints of $[-1,1]$ is to note the equality $$int_{0}^{x}sqrt{1-t^2},dt=frac{xsqrt{1-x^2}}{2}+frac{1}{2}int_{0}^{x}frac{dt}{sqrt{1-t^2}}$$ valid for $xin(-1,1)$. Taking limits as $xto pm 1$ shows that the integral in question converges if $x=pm 1$.
– Paramanand Singh
Nov 30 at 17:19
add a comment |
This is good for the open interval $(-1,1)$. However, you need to show the endpoints separately. Consider $g(x) = frac{1}{1-|x|}$ continuous on $(-1,1)$ but $int_0^x g(t)text dt$ not defined for $x in {-1,1}$.
– AlexanderJ93
Nov 30 at 6:29
1
@AlexanderJ93 Edited my post.
– Guacho Perez
Nov 30 at 7:00
Another approach to deal with endpoints of $[-1,1]$ is to note the equality $$int_{0}^{x}sqrt{1-t^2},dt=frac{xsqrt{1-x^2}}{2}+frac{1}{2}int_{0}^{x}frac{dt}{sqrt{1-t^2}}$$ valid for $xin(-1,1)$. Taking limits as $xto pm 1$ shows that the integral in question converges if $x=pm 1$.
– Paramanand Singh
Nov 30 at 17:19
This is good for the open interval $(-1,1)$. However, you need to show the endpoints separately. Consider $g(x) = frac{1}{1-|x|}$ continuous on $(-1,1)$ but $int_0^x g(t)text dt$ not defined for $x in {-1,1}$.
– AlexanderJ93
Nov 30 at 6:29
This is good for the open interval $(-1,1)$. However, you need to show the endpoints separately. Consider $g(x) = frac{1}{1-|x|}$ continuous on $(-1,1)$ but $int_0^x g(t)text dt$ not defined for $x in {-1,1}$.
– AlexanderJ93
Nov 30 at 6:29
1
1
@AlexanderJ93 Edited my post.
– Guacho Perez
Nov 30 at 7:00
@AlexanderJ93 Edited my post.
– Guacho Perez
Nov 30 at 7:00
Another approach to deal with endpoints of $[-1,1]$ is to note the equality $$int_{0}^{x}sqrt{1-t^2},dt=frac{xsqrt{1-x^2}}{2}+frac{1}{2}int_{0}^{x}frac{dt}{sqrt{1-t^2}}$$ valid for $xin(-1,1)$. Taking limits as $xto pm 1$ shows that the integral in question converges if $x=pm 1$.
– Paramanand Singh
Nov 30 at 17:19
Another approach to deal with endpoints of $[-1,1]$ is to note the equality $$int_{0}^{x}sqrt{1-t^2},dt=frac{xsqrt{1-x^2}}{2}+frac{1}{2}int_{0}^{x}frac{dt}{sqrt{1-t^2}}$$ valid for $xin(-1,1)$. Taking limits as $xto pm 1$ shows that the integral in question converges if $x=pm 1$.
– Paramanand Singh
Nov 30 at 17:19
add a comment |
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continuous functions are Riemann integrable.
– Lord Shark the Unknown
Nov 30 at 5:40
How would one construct the epsilon-delta continuity proof for that function, though?
– MathIsLife12
Nov 30 at 5:43
Are you just trying to prove that $f(x) = int_0^x frac{1}{sqrt{1-t^2}}text dt$ is a continuous odd function on the interval $[-1,1]$?
– AlexanderJ93
Nov 30 at 5:49
1
@AlexanderJ93 Yes. The question is worded weirdly but that is what it comes down to. I am a little stuck.
– MathIsLife12
Nov 30 at 5:51
You are on the right track, just let $t to -t$ in the last integral and simplify.
– Mustafa Said
Nov 30 at 6:02