Find inverse hyperbolic function

In summary, the solution to finding f-1(t) and computing f-1(0) where f(t) = sinh(t) is to first solve for y(t) = sinh(t) by setting y = (et- e-t)/2 and using the quadratic formula to find the value of t. Then, let x = et and solve for t(y) by taking the logarithm of both sides. Finally, plug in t = ln(e^y) = y into the equation f-1(t) = t-\sqrt{t^2+1} to find the inverse function and f-1(0) = ln1 = 0.
  • #1
synergix
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0

Homework Statement


find f-1(t) and compute f-1(0) where
f(t) = sinh(t)

Homework Equations


sinh(t)=( et- e-t)/2

The Attempt at a Solution



t= (ey- e-y)/2

ln2t = y - - y

y = (ln2t)/2

but wikipedia says
http://upload.wikimedia.org/math/8/e/d/8edd21edb4cd413e462fe82ef4ac249b.png"

what silly mistake have i made this time?
 
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  • #2
i would check the step where you take the logarithm

ony way that may work, would be to let y = f(t), x=e^t then try and solve for x(y) first
 
  • #3
y = (et- e-t)/2
2y = et- e-t
2yet = e2t - 1 (multiply both sides by et)

Let x = et, then you have a quadratic equation to help you isolate the t.
 
  • #4
yes! this is how my instructor showed me but we have covered soo much this past couple of weeks I completely forgot about that technique. Thanks!
 
  • #5
I used z instead of x and I got to
z^2-2tz-1=0
z= t+- sqrt(t^2+1)
z = e^y

so ln both sides

how do i know whether it is positive or negative?

since i can see from the formulas on wikipedia I went ahead with the rest and f-1(0) = ln1 = 0
 
Last edited:
  • #6
so is [itex] t-\sqrt{t^2+1} [/itex] greater or less than zero? is this for all t? consider what happens when you take the logarithm, and which is actually a solution to the orginal equation

also, I find the way you swap y & t is quite confusing
you start with:
[itex] y(t) = sinh(t) [/itex]

so if I was you, I would solve for t(y)
 

Related to Find inverse hyperbolic function

What is an inverse hyperbolic function?

An inverse hyperbolic function is a mathematical operation that undoes the effect of a hyperbolic function. It is denoted as "arcsinh", "arccosh", "arctanh", "arccsch", "arcsech", or "arccoth", depending on the specific hyperbolic function being inverted.

Why is finding the inverse hyperbolic function important?

Finding the inverse hyperbolic function is important because it allows us to solve equations involving hyperbolic functions. It also helps in simplifying complex expressions and solving problems in various fields such as physics, engineering, and mathematics.

How do you find the inverse hyperbolic function?

To find the inverse hyperbolic function, we use the inverse hyperbolic trigonometric identities. These identities can be found in any standard trigonometry or calculus textbook. We also use algebraic manipulation and substitution to simplify the expression and solve for the inverse function.

What are some real-world applications of inverse hyperbolic functions?

Inverse hyperbolic functions have many real-world applications, such as in physics for solving problems involving motion and acceleration, in engineering for designing curves and surfaces, and in statistics for analyzing data distributions. They are also used in finance, biology, and other fields where exponential growth or decay is involved.

Are there any limitations to finding the inverse hyperbolic function?

Yes, there are some limitations to finding the inverse hyperbolic function. One limitation is that not all hyperbolic functions have an inverse function that can be expressed in terms of elementary functions. In these cases, we may have to use numerical methods to approximate the inverse function. Additionally, finding the inverse hyperbolic function can be challenging and time-consuming for complex expressions, so it is important to have a good understanding of trigonometry and algebraic manipulation techniques.

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