Prove that $mathrm{arcsinh}(x)$ is an odd function
The inverse hyperbolic sine $sin^{-1}(x) = mathrm{arcsinh}(x)$ is an odd function. This can be proved by manipulating the expression $mathrm{arcsinh}(-x) = y$ as shown here.
But how to prove it through this definition instead?
$$mathrm{arcsinh}(x) = log left( x + sqrt{x^2 + 1} right)$$
My attempt:
I applied this rule, so obtaining
$$mathrm{arcsinh}(-x) = log left( sqrt{x^2 + 1} right) + log left( 1 - frac{x}{sqrt{x^2 + 1}} right)$$
This does not seem the expression of $mathrm{arcsinh}(x)$ with opposite sign. How to proceed?
calculus real-analysis logarithms hyperbolic-functions
asked Nov 24 at 14:34
BowPark
549720
add a comment |
The inverse hyperbolic sine $sin^{-1}(x) = mathrm{arcsinh}(x)$ is an odd function. This can be proved by manipulating the expression $mathrm{arcsinh}(-x) = y$ as shown here.
But how to prove it through this definition instead?
$$mathrm{arcsinh}(x) = log left( x + sqrt{x^2 + 1} right)$$
My attempt:
I applied this rule, so obtaining
$$mathrm{arcsinh}(-x) = log left( sqrt{x^2 + 1} right) + log left( 1 - frac{x}{sqrt{x^2 + 1}} right)$$
This does not seem the expression of $mathrm{arcsinh}(x)$ with opposite sign. How to proceed?
calculus real-analysis logarithms hyperbolic-functions
asked Nov 24 at 14:34
BowPark
549720
2
$(x+sqrt{x^2+1})(-x+sqrt{x^2+1})=1$
– Lord Shark the Unknown
Nov 24 at 14:37
2
The rule pulls you away from the solution, don't use it.
– Yves Daoust
Nov 24 at 14:39
add a comment |
The inverse hyperbolic sine $sin^{-1}(x) = mathrm{arcsinh}(x)$ is an odd function. This can be proved by manipulating the expression $mathrm{arcsinh}(-x) = y$ as shown here.
But how to prove it through this definition instead?
$$mathrm{arcsinh}(x) = log left( x + sqrt{x^2 + 1} right)$$
My attempt:
I applied this rule, so obtaining
$$mathrm{arcsinh}(-x) = log left( sqrt{x^2 + 1} right) + log left( 1 - frac{x}{sqrt{x^2 + 1}} right)$$
This does not seem the expression of $mathrm{arcsinh}(x)$ with opposite sign. How to proceed?
calculus real-analysis logarithms hyperbolic-functions
asked Nov 24 at 14:34
BowPark
549720
The inverse hyperbolic sine $sin^{-1}(x) = mathrm{arcsinh}(x)$ is an odd function. This can be proved by manipulating the expression $mathrm{arcsinh}(-x) = y$ as shown here.
But how to prove it through this definition instead?
$$mathrm{arcsinh}(x) = log left( x + sqrt{x^2 + 1} right)$$
My attempt:
I applied this rule, so obtaining
$$mathrm{arcsinh}(-x) = log left( sqrt{x^2 + 1} right) + log left( 1 - frac{x}{sqrt{x^2 + 1}} right)$$
This does not seem the expression of $mathrm{arcsinh}(x)$ with opposite sign. How to proceed?
calculus real-analysis logarithms hyperbolic-functions
calculus real-analysis logarithms hyperbolic-functions
asked Nov 24 at 14:34
BowPark
549720
asked Nov 24 at 14:34
BowPark
549720
asked Nov 24 at 14:34
BowPark
549720
asked Nov 24 at 14:34
BowPark
549720
asked Nov 24 at 14:34
BowPark
549720
549720
2
$(x+sqrt{x^2+1})(-x+sqrt{x^2+1})=1$
– Lord Shark the Unknown
Nov 24 at 14:37
2
The rule pulls you away from the solution, don't use it.
– Yves Daoust
Nov 24 at 14:39
add a comment |
2
$(x+sqrt{x^2+1})(-x+sqrt{x^2+1})=1$
– Lord Shark the Unknown
Nov 24 at 14:37
2
The rule pulls you away from the solution, don't use it.
– Yves Daoust
Nov 24 at 14:39
2
2
$(x+sqrt{x^2+1})(-x+sqrt{x^2+1})=1$
– Lord Shark the Unknown
Nov 24 at 14:37
$(x+sqrt{x^2+1})(-x+sqrt{x^2+1})=1$
– Lord Shark the Unknown
Nov 24 at 14:37
2
2
The rule pulls you away from the solution, don't use it.
– Yves Daoust
Nov 24 at 14:39
The rule pulls you away from the solution, don't use it.
– Yves Daoust
Nov 24 at 14:39
add a comment |
1 Answer
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$$text{arsinh}(x)+text{arsinh}(-x)=log(sqrt{x^2+1}+x)+log(sqrt{x^2+1}-x)=log(x^2+1-x^2).$$
You can conclude.
answered Nov 24 at 14:38
Yves Daoust
123k670220
add a comment |
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$$text{arsinh}(x)+text{arsinh}(-x)=log(sqrt{x^2+1}+x)+log(sqrt{x^2+1}-x)=log(x^2+1-x^2).$$
You can conclude.
answered Nov 24 at 14:38
Yves Daoust
123k670220
add a comment |
$$text{arsinh}(x)+text{arsinh}(-x)=log(sqrt{x^2+1}+x)+log(sqrt{x^2+1}-x)=log(x^2+1-x^2).$$
You can conclude.
answered Nov 24 at 14:38
Yves Daoust
123k670220
add a comment |
$$text{arsinh}(x)+text{arsinh}(-x)=log(sqrt{x^2+1}+x)+log(sqrt{x^2+1}-x)=log(x^2+1-x^2).$$
You can conclude.
answered Nov 24 at 14:38
Yves Daoust
123k670220
$$text{arsinh}(x)+text{arsinh}(-x)=log(sqrt{x^2+1}+x)+log(sqrt{x^2+1}-x)=log(x^2+1-x^2).$$
You can conclude.
answered Nov 24 at 14:38
Yves Daoust
123k670220
answered Nov 24 at 14:38
Yves Daoust
123k670220
answered Nov 24 at 14:38
Yves Daoust
123k670220
answered Nov 24 at 14:38
Yves Daoust
123k670220
123k670220
add a comment |
add a comment |
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2
$(x+sqrt{x^2+1})(-x+sqrt{x^2+1})=1$
– Lord Shark the Unknown
Nov 24 at 14:37
2
The rule pulls you away from the solution, don't use it.
– Yves Daoust
Nov 24 at 14:39