Can /b help?

can /b help?
I need help solving for x, step by step if possible
I know that x can be 0 or approximately 0.76

Don't know if it's possible to get an exact solution.

1. Cut your wrists off
2. Cut your dick off
3. Drink a shit ton of alcohol
4. templatestatic std::unique_ptr CreUniquePtr(T& type, Args... args);

don't need the exact solution, just the way to get an approximate solution
eat a dick, nigger

drink some bleack faggot

Newton's approximation around 0 and 1 should converge to your two solutions

>drink some bleack faggot
>some bleack
>bleack
having a hard time writing with your down syndrome, fag?

>graph both sides
>find intersection
2EZ

you could use the taylor series for cosh if you only need a approximately solution.

hope this helps, i havent started working with the term "cosh" yet so i dont know any legit ways of doing this.

so 0 is my x0?
Wolfram Alpha gave me x ≈ 0.76, so I guess there isn't a solution without ∞ digits after 0.7.....

Y(x) =2/5 x+1
If x=o
Y(0)=2/5 (0)+1 answer: 1
If x=.76
Y(.76)=2/5(.76)+1

So find what 2/5 of .76 is then add 1.

Aint rocket science. Its algabra

TIENES QUE GRAFICAR AMBAS EXPRESIONES.

this man knows what's up

you do an approximation for every solution you want. So in theory you need to find out how many solutions there are, then you can use bisections to find appropriate values that will converge. In this case I would use x0 = 0 and x0 = 1, and you should arrive at your two solutions

that would probably be easiest

Thanks Mr. Obvious

What a nigger

Is there some way of simplifying this into an exact function result for that approximately 0.76 answer? Because I don't see how...

Hyperbolic cosine.

maybe it'll help if I tell you it's the formula for getting the horizontal pull H for a catenary and x is q*l/2H with given q and l

No there is not. You can approximate the solutions by either approximating cosh(x) with it's taylor series, or approximating the equation with newton's approximation, or you can get your crayons out and draw the graphs like a child

mathematician here,
the solutions (for reell x) are x=0 and x=~1.61614
I don't think it's possible to write down the second solution exactly. So I can't give you a "step py step" solution. What exactly do you need OP?

just use the approximation with the taylor series, you have enough other factors influencing your actual solution.

google shit like that, annon..

dont b them..

gnomewotimeen?

been years since i finished uni..dont remember..

Use google as: blah blah blah - case study. case study will help u a lot in finding answers, or even complete courseworks, similar ot your topic.

u are welcome young fags.

(civil engineer,became 32 today)

see I need x to solve for H

happy birthday, but this part is the only problem I have, and all I get on the net are the solutions, but I need the path towards it

>x=~1.61614
Oh I forgott the 2/5. Of xourse the actual solutions are x=0 and x=~0.762355.

jesus fuck dude, we already give you the way. Either use taylor series or newton's approximation

thanks.. thats cool..dont downplay the 'case study' tip..

mathematician fag needs to make a pic of his solution and post it here..

it aint rocket science ffs..

I don't give a fuck nigger

even tho i dont really remember, since im an oldfag as ive already stated..

ill try to help, my friend.

cosh x = (e^x + e^-x)/2

then u probably differentiate this equation... since e^x is cool with that..

this could be a step..

so now u can take the e^x part and differentiate both sides... 2/5x...x goes away.. and u get something..

work with it annon..

ah, now I get it
I kept forgetting that cosh(x)=1/2(e^x+e^-x)
still would have liked mathfag's solution

Mathfag here
Will post taylor-solution, just keep thread alive

awesome, thank you

>not giving a fuck?

my approach was sweet as well.. made that little boy happy..

cosh(x)=1/2(e^x+e^-x)

glad i could help.. havent done Taylor since '05-'06

holy shit, how about you use your own brain for 5 seconds.

first degree taylor series for cosh(x) = 1 + x^2/2

=> 1 + x^2/2 = 2/5 x + 1
=> x1 = 0
=> x/2 = 2/5
=> x2 = 4/5

if you want better solutions use higher degree taylor approximation and solve for x

lets keep this thread alive for the sake of op.


its a feminine penis thread till we get the solution..

i dont have any.. so lets rooooll

your approach doesn't do shit faggot, keep play pretending that civil engineers are actual engineers and let the adults do the math

come on, he meant well, at least he didn't wish for anybody to drink bleach

no it isn't, fag

sorry for not washing my balls today..

u do seem to enjoy yourself swinging up and down my balls and cock though.. ..

>i just feel like apologising for the 'aroma'

sit on my cock u jelous peasant

bbuut they are chicks..they just have a penis..

>its feminine

thinking is not fun.> j/k, >lol,rofl,kekathlon..
>errm..adjusts tie, wipes fedora of sweat.. ..
>goes back to studying

>samefagging that hard

go cry some more fag

nigger

fuck off

If approximate solution is ok, you can use a series expansion. Since

cosh(x) = (e^x - e^(-x)) / 2

the series expansion is

1 + x^2/2 + x^4/24 + ...

This converges fairly rapidly even at x = 0.76; x^4/24 is about 0.014 when x = 0.76. So we can approximate this as

1 + x^2/2 = 2/5 x + 1

x/2 (x - 4/5) = 0

which gives x = 0 or x = 0.8. Not too bad. If you kept the next term in the expansion you'd have a cubic to solve but would get a more accurate answer.

Here we go.
If you choose a higher grade of the taylor series you get better aproximations, but it becomes much more difficult to solve (polynomes of order 4 and above). So then in the end you would also end up Newton approximation. But newton approximation is tedious. That's the reason why We use calculators and computers for such shit nowadays.

X=0

thank you very much, I will look into it later

you're welcome

Your post isn't a solution at all. It's just an alternative notation