Sorta stuck help would be appreciate

Well I need to solve this equation for x but I am baffled

0.5=(e^0.6x + e^-0.6x)/(e^x+e^-x)

Thanks

The substitution y=e^0.4x transforms this into a quintic, of which (y+1) is a spottable factor (the solution y=-1 is extraneous, though). After division, there is a quartic which could be solved by the quartic formula. If this is the expected avenue of solution, it wouldn't seem a very good problem, so perhaps there's a better way.

Reply 2

nuodai

Are you meant to find the exact value of

x

or some approximation to it (for example, using numerical methods)?

In any case, you can write the RHS as

\dfrac{\cosh \frac{3x}{5}}{\cosh x}

. Alternatively, you can substitute

u=e^{0.2x}

and rearrange to obtain the equation

u^5-2u^3-2u^{-3}+u^{-5}=0

, which isn't too hard to solve numerically.

Reply 3

PeanutButterJellyTime

Original post by nuodai

Are you meant to find the exact value of

x

or some approximation to it (for example, using numerical methods)?

In any case, you can write the RHS as

\dfrac{\cosh \frac{3x}{5}}{\cosh x}

. Alternatively, you can substitute

u=e^{0.2x}

and rearrange to obtain the equation

u^5-2u^3-2u^{-3}+u^{-5}=0

, which isn't too hard to solve numerically.

Just tells to estimate x

Reply 4

nuodai

Original post by PeanutButterJellyTime

Just tells to estimate x

Right; in that case a numerical method like iteration would be best. It's quite easy to rearrange one of the expressions I gave above so that you get an iterative formula.

Reply 5

PeanutButterJellyTime

The question is Ca(y)=cosh m(L-y)/cosh(ML) tells me to take x=ml and Ca(y)=0.5 and y=0.4and the cosh relation ship

Reply 6

PeanutButterJellyTime

Original post by nuodai

Right; in that case a numerical method like iteration would be best. It's quite easy to rearrange one of the expressions I gave above so that you get an iterative formula.

This might be a dumb question but why do you make it u=e^0.2?

Reply 7

nuodai

Original post by PeanutButterJellyTime

This might be a dumb question but why do you make it u=e^0.2?

Because 2 is the highest common factor of 6 and 10, which meant that both

e^{\pm 0.6x}

and

e^{\pm x}

could be expressed as integer powers of

e^{0.2 x}

. It just simplifies notation.

Reply 8

Mr M

Study Forum Helper

Original post by nuodai

Right; in that case a numerical method like iteration would be best. It's quite easy to rearrange one of the expressions I gave above so that you get an iterative formula.

You don't need to iterate.

t^5-2t^4-2t+1=(t+1)(t^4-3t^3+3t^2-3t+1)

Now deal with the quartic.

Divide through by

t^2

and make the substitution

w=t+\frac{1}{t}

Now solve the quadratic in

w

.

The two real answers are horribly long logarithms with nested surds but it is quite satisfying to do.

Edit: Just seen that only an estimate is required!

(edited 12 years ago)

Reply 9

MrGum

Original post by Mr M

You don't need to iterate.

t^5-2t^4-2t+1=(t+1)(t^4-3t^3+3t^2-3t+1)

Now deal with the quartic.

Divide through by

t^2

and make the substitution

w=t+\frac{1}{t}

Now solve the quadratic in

w

.

The two real answers are horribly long logarithms with nested surds but it is quite satisfying to do.

Edit: Just seen that only an estimate is required!

I thought there had to be a better route than the quartic formula. Thanks for pointing it out!