If you can't do this, you need to go back to grade school

>>9042576
http://www.wolframalpha.com/input/?i=sin(1%2Bx)+%3D+x

>>9042578
>he can't express x in closed form or as an infinite sum

lol It's like you were dropped on your head as a baby.

>>9042576
sin(x) = x
therefore x = 1+sin(1+sin(1+sin(1+...

>>9042583
http://www.wolframalpha.com/input/?i=faggot

File: Problem 4 u.jpg (16KB, 350x200px) Image search: [Google]

16KB, 350x200px

More fun:

[eqn] ln(a+ln(a+ln(a+ln(a+... = x [/eqn]

[eqn]x^{x^{x^x...}} = 6[/eqn]

That's x^x repeating all the way to infinity.

And finally, pic related. Bonus: find the solution for ALL values of a and b

~0.934

>>9042576
x=sin(1+x)

0.934563

>>9042621
For the second one: [eqn] \text{log}_x(6)=6 \\
\frac{\text{ln}(6)}{\text{ln}(x)}=6 \\
\text{ln}(x)=\frac{\text{ln}(6)}{6} \\
x=(e^{\text{ln}(6)})^{\frac{1}{6}} \\
x=6^{\frac{1}{6}}\approx 1.35 [/eqn]

>>9042657
Not so fast, this problems fights back.

Why don't you plug that in and see what you get? :^)

>>9042690
6

>>9042700
Brainlet.

>>9042700
Bruh.

Put your answer "1.35...." into the calculator, raise it to the power of "1.35..." and keep doing this. Keep doing this over and over again and see what you get.

>>9042621
a=91x/9
b=x^2

>>9042795
>a=91x/9
>91x/9
>x
>what the fuck am i reading.jpg

Limit as x approaches zero, bro. Find a value for a and b so that it converges to 1/3 when x approaches 0.

>>9042739
Sorry, I don't have time to perform an infinite number of calculations. I'll just stick to logic and reasoning: [eqn] 6^{\frac{1}{6}^{6^{\frac{1}{6}^{6^{\frac{1}{6}^{\cdot^{\cdot^\cdot}}}}}}}=6^{1^{1^{1^\cdots}}} [/eqn]
Basic properties of powers

>>9042801
no restrictions on a and b in the problem
unless I have made an algebra mistake

File: 1486095430786.png (511KB, 836x964px) Image search: [Google]

511KB, 836x964px

>>9042804

>>9042804
>Sorry, I don't have time to perform an infinite number of calculations
Retard? Just do it 10 times and you'll see it goes way beyond six.

>Basic properties of powers
I like how you give this hand-waving argument without actually showing the property. Hint: because it doesn't exist.

According to you:

[eqn]6^{\frac{1}{6}^{6^{\frac{1}{6}^{6^{\frac{1}{6}^{\cdot^{\cdot^\cdot}}}}}}}=6^{1^{1^{1^\cdots}}}[/eqn]

Ok, Let's raise both sides with base [math]6^\frac{1}{6}[/math]^

[eqn]6^{\frac{1}{6}^{6^{\frac{1}{6}^{6^{\frac{1}{6}^{\cdot^{\cdot^\cdot}}}}}}}= 6^{\frac{1}{6}6^{1^{1^{1^\cdots}}}}[/eqn]

Basic properties of powers:

[eqn]=6^\frac{1}{36}[/eqn]

Does this shit look the same to you?

[eqn]6 = 6^\frac{1}{36}?[/eqn]

This is BASIC, BASIC shit. If you can figure this out you should probably be wearing a helmet when you go outside.

>>9042826
What kind of crack are you smoking and how can I get my hands on some?

>>9042807
Dude, plug in x = 0 and you will get:

a = 0, b = 0

Plug this into the original and you get

[eqn]\frac{\sqrt{(0)x + 0}-3}{x}[/eqn]

Which is

[eqn]\frac{-3}{x}[/eqn]

That does not converge as x goes to zero, that blows up to infinity or negative infinity. If you don't know what I'm talking about read this:

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/limcondirectory/LimitConstant.html

>>9042859
just misread -3 as being outside the fraction

>plug in x = 0
wew

>>9042873
Read the link, little man.

>>9042879
>plug in x = 0
>
this is an example of poor teaching
giving a method instead of reason
this also may mislead students further down the line if they use this without thought

>consider when x is very small, the expression will be close to -3/x
gives much more insight, littler man

>>9042893
In the step I plug in x=0 I'm evaluating a and b, not the entire function you retard.

[eqn]\lim_{x \to 0 } a = 0[/eqn]

[eqn]\lim_{x \to 0 } b = 0[/eqn]

Now that we know the values of a and b we can plug this into the function and see that you're wrong.

The fact that you are stalling so hard proves you don't know how to solve this. Sad! Any pre-calc student can solve this shit easily. But you struggle very hard.

Of course if you gave me the solution I would have to eat my words, but I think I'll be going hungry tonight.

Write [math]x \,=\, \sin\left( 1 \,+\, x\right)[/math], then use Euler's method (convergence conditions are met almost everywhere). Happy homework!

>>9042907
I have admitted my mistake
>>9042873
>just misread -3 as being outside the fraction

lets see if you can admit yours

let a=x
now I will use your method to evaluate lim x->0 of x/(a+x)
lim x->0 a = 0
so now x/(a+x) = x/(0+x) = x/x
lim x->0 x/x = 1

but if we sub in a=x earlier we get x/(a+x) = x/2x
then lim x->0 x/2x = 1/2

1=1/2

whoah

>>9042933
We'll there's no hiding that yeah, I fucked up on that one.

>>9042932
How the fuck would you use Eulers method here?

>>9042982
he prob means newton's method

0.9345632108

>>9042826
Hey man this is wrong

(a^b)^c =/= a^(b^c)

>>9042576
I actually have no clue how to do this
Where can I read about it?

>>9042826
>reddiot spacing
>is an idiot
like pottery
at least you just left the thread instead of deleting your post

>>9043192
recognize that the series is recursive, and use symbolic replacement. it's not really something you need to focus to learn how to do.

[math] \sin(1 + \sin(1 + \sin(1 + ... = x [/math]
now look at it again
[math] \sin(1 + (\sin(1 + \sin(1+ \sin(1 + ... )))) = x [/math]
see it?
we can replace that infinite recursion with what we had in the first place
[math] \sin(1+x) = x [/math]
this you should be able to solve

>>9043192
Honestly watch this video:

https://www.youtube.com/watch?v=leFep9yt3JY

One other thing I've noticed is with these type of things, you can always see the decimals end the exact same.

Take the golden ratio, which can be written as an infinite continued fraction of 1s, if you take:

[eqn](1.6180339...)^2 = 2.6180339...[/eqn]

Or look at this number:

[eqn]1.776775...^{1.776775...} = 2.776775...[/eqn]

Or this:

[eqn]sin(1.934563...) = 0.934563...[/eqn]

Something special about these numbers man..

>>9043222
>this you should be able to solve
Go ahead and solve for x, I dare you.

There is no closed form solution.

>>9042576

how is it a well-formed expression without a base case?

>>9043228
then use a machine.

awfully mean to trick brainlets with this, /sci/

>>9042621
Obviously, a=2 and b=9

>>9042576
By the Banach fixed-point theorem, there exists an [math]x^*[/math] such that [math]sin(1 + x^*) = x^*[/math]:
Define [math]x_{k+1} := sin(1+x_k)[/math].
Let [math]x^* := \lim_{k \rightarrow \infty} x_{k+1}[/math].
Proving this limit exists is left as an exercise to the reader.
Then, since [math]sin(1+x)[/math] is continuous, [math]sin(1 + x^*) = x^*[/math].
It turns out that [math]x^* \approx 0.9345632108[/math]

>>9043421
>Proving this limit exists is left as an exercise to the reader.
meme master

>>9043421
The limit exists by the Banach fixed-point theorem.

>>9043761
The proof for the theorem uses exactly the fact that this iterative sequence converges.

>>9043421
couldnt you just do IVT on f(x) = sin(1+x) - x

>>9043845
Sure, but this is a constructive proof.