Finding x here

Calculus, Early Trascendentals - 6th edition.
Page 181 excercise 48

because that way I cand find a maximum and minimums and then I can graphic

ur a dumb ass
f(1) = e-3 0
so theres a root in there

You can sketch a function without exact forms for the max/min points

Just went through Newton's iterative method and got x ~= 0.9100
There is probably at least one more root in there though.

it is a transcendant equation, no analytic solutions exist. Just like x=tan(x)

but I don't want to use a calculator
I want to do it on my own.because in the tests and exams I can't use one

non-elementary, might be a lambert W function case, but idk. If it's from stewart than I'm almost sure it's a numerical methods question, newton's method as someone mentioned

OP here, I just used Wolfram Alpha to solve it and it showed me the W thing but I don't even know that newton's method, I don't have it in my class yet

sure buddy

google.com/#q=y=e^x-3*x*x

How do you find x? You don't.
There's no analytical solution.
The best geometric way to do that would be to use Newtons me

[math] 0 = f(x0) = f'(x0)(x1-x0) [/math]
so
[math] x1 - x0 = - \frac {f(x0)}{f'x0} [/math]
=>
[math] x1 = x0 - \frac {f(x0)}{f'x0} [/math]

**Newtons Method

x0 is any number you want to start with, guestimate this, the closer the guess the less approximations.

Find the tangent line to f(x) at x0
Where f'(x0) crosses the x axis, that is x1
x1 is now your new x0. Do it again. Continue until the numbers you plug back into f(x) =~0

This is all a computer does to calculate solutions.

google.com/#q=y=e^x-3*x*x,e^x,-3*x*x

[math] x = \frac{1}{\sqrt{3}} e ^ {\frac{1}{\sqrt{12}} e ^ {\frac{1}{\sqrt{48}} ...}} [/math]

This, but have fun evaluating that.
>inb4 this is the exact answer
His book probably wants a decimal answer to 2-3 digits, remember that calc is literally engineer tier plug and chug at this point :^(
>T. EE

OP here.
My teacher didn't teach me that, But it's similar to the tangent line formula

[math] x = \ln(3) + 2\ln(\ln(3) + 2\ln(\ln(3) + 2\ln(...))) [/math]

This guy manages to convey the idea in about 1 minute

youtu.be/1uN8cBGVpfs?t=30s

In fact it's the same thing, that's what's so exciting.
[math] y = f(a) + m (x - a) [/math]
Replace x with x1 and a with x0, and m with f'(x)
y = 0 since that's literally what your equation says (and we know that happens at some point Xn,0)
[math] 0 = f(x0) = f'(x0) (x1 - x0) [/math]

It's literally using the tangent line to get very very close to the interception point, as when you get closer graphically to the point of interception, your tangent line at that point will also intercept the x axis very close to the graphical interception of your function on the x axis.

***
[math] 0 = f(x0) + f'(x0) (x1 - x0) [/math]

Woah, I was freaking out why [math] e^x - 3x^2 =0 [/math] had three solutions on desmos, but
[math] e^x = 3x^2 [/math] only had two...

Then I zoomed out. Ecks dee.

But yeah, OP if you *didn't* learn Newton's method and this isn't a real analysis class then there'd be no way to do the problem.

>here's no analytical solution.
Meant algebraic
There's obviously an analytical solution.

Here's how the problem looks in my book. The graphing symbol next to the problem number means you're suppose to use outside graphing software, not do it by hand. The problem is asking you to make sure to formula you compute is consistent with how the graph looks.

cute

had to think about it for a second

x solves the given equation iff it is a fixed point of the function [math] f(x)=e^{x/2}/\sqrt{3}[/math]

or of the function

[math] g(x)=-e^{x/2}/\sqrt{3}[/math]

the mean value theorem implies that f and g are contraction mappings on any interval [math] (-\infty, a] [/math] where f'(a)

just zoomed out on the graph, and there's actually a third fixed point that you won't find with this method (it's approximately x=3.7>2ln(2\sqrt(3)), which means that f and g are not contraction mappings in a neighborhood of the point)

This would be from Calc I. What do you mean 'already plug and chug at this point?' What does that make the rest of the Calcs? And what would you prefer to see Calc taught as? Not arguing w/ you, just genuinely curious what you mean as I also feel the majority of college math is 'plug n chug.'

OP here
I actually find that fascinating
thank you for that video, I speak spanish but i get those terms
I think i just learnt it
I didn't know the book implied the use of a software
Still dont get that
What do you mean with 'plug n chug'
-------
Thank you everyone!! I think I'm improving here.

Use Newton's method. You know you can't solve any arbitrary degree 5 polynominal equation in terms of radicals, so why do you think you are going to solve e^x - 3x^2 = 0 with a single sheet of paper? What kind of solution do you think can be written on that paper? If you choose decimal approximation, then your best bet is Newton's method.

You're stupid. e^x = 1 when x = 0, and 3x^2 = 0 when x = 0. However, when x = 1, we have it that e^x = e < 3, but 3x^2 = 3. Using the intermediate value theorem, one can easily prove that e^x - 3x^2 = 0 for some x on (0,1).

Integrate at C=1 x=0 f(x)=1.

The first thing you should have done is to STUDY the function exp(x)-3x^2.

You don't need to use special functions, you can first find the number of solutions. It's already a big thing. Then you can find approximations for these solutions.

You derive it once : exp(x)-6x
You derive it twice : exp(x)-6
You find variations, and you conclude thanks to the variation chart of your initial function. This is how it should be done properly.

Why not the exact solution using Lambert W function (W(z) is the solution to z=W*exp[W], see for example press.princeton.edu/chapters/s3-17_10592.pdf)?
x=-2W(1/(2*sqrt[3]))
And you only have to look it in a table or use one of the expansion of W.

close but no cigar; you want to compare e^x and 3x^2 and find the intersections.
you just provided an analysis of the equation e^x + 3x^2 = 0 for which there is no solution.

I'm an engineer, I solve problems like this pretty much every day in my line of work.

Is that calc 1? I forget.

Either way the way the calculus classes are taught, ie differentiation (implicit, explicit, partial), integration, integration techniques and applications etc etc all throughout the calc 1-3's is extremely plug and chug. Overwhelmingly I've found that the only difference between a physics class and a math class is that sometimes the Physics class hands you a formula sheet.

Derivatives are literally plug and chug
D/dx X^n = nX^n-1..
D/dx F(g(x)) =F'(g(x))*g'(x)
"Quotient rule", "product rule".
I don't expect calc 1 studenta to derive it for themselevs but other than in math major specific calc and analysis classes I've never seen derivation or integration explained other than "rate of change and here's the formula" or "summing area / anti derivative and here's the formula".

...

e^x=3x^2
ln(e^x)=x
ln(3x^2) = ln(3) + ln(x^2)
Isolate, subtract, root

e^x - 3x^2 = 0

(e^x/2 - sqrt3 x) (e^x/2 sqrt3 x) = 0

(e^x/2 - sqrt3 x) = 0 or (e^x/2 sqrt3 x) = 0

Which requires the product log function

So what kind of equations can the Lambert function be used to solve. So far I've got
y^(a x^b) = (c x^d)^(e x^f)
y^(a/e x^b) = (c x^d)^(x^f)
y^(a/e x^(b/f)) = (c x^(d/f))^x
y^(a/e x^(b/f-1)) = c x^(d/f)
y^(a/e x) = c x^(d/(b-f))
y^(a/e (b-f)/d x) = c^((b-f)/d) x
1 = c^((b-f)/d) x y^(a/e (f-b)/d x)
a/e (f-b)/d c^((f-b)/d) = a/e (f-b)/d x y^(a/e (f-b)/d x)
a/e (f-b)/d log(y) c^((f-b)/d) = a/e (f-b)/d log(y) x exp(a/e (f-b)/d log(y) x)
a/e (f-b)/d log(y) x = W(a/e (f-b)/d log(y) c^((f-b)/d))
x = W(a/e (f-b)/d log(y) c^((f-b)/d))/(a/e (f-b)/d log(y))

there are no real solutions OP

must be a misprint

...

Are you familiar with the product log function?
wrong btw

engineers everyone

0

[math]
e^x - 3x^2 = 0
e^x = 3x^2
d/dx (e^x) = d/dx (3x^2)
e^x = 6x
d/dx (e^x) = d/dx (6x)
e^x = 6
ln (e^x) = ln 6
x = ln 6 =1.79
[/math]

Am I the only smart one in this thread? lol you guys are fucking stupid.

*reformatting*

[math]e^x - 3x^2 = 0[/math]
[math]e^x = 3x^2[/math]
[math]d/dx (e^x) = d/dx (3x^2)[/math]
[math]e^x = 6x[/math]
[math]d/dx (e^x) = d/dx (6x)[/math]
[math]e^x = 6[/math]
[math]ln (e^x) = ln 6[/math]
[math]x = ln 6 =1.79[/math]

Am I the only smart one in this thread? lol you guys are fucking stupid.

dont you do this pre-calc
geometric series or something.

>i never graduated highschool so i dont remember
>im only here to shitpost about global warming

Isn't this correct? I don't understand why this isn't right

if you plug in ln6 back into the equation it doesn't satisfy the initial condition you fucking traffic cone

0.91 works

can confirm this works

How do you just cross x and 2 out in step 2?

x = 0.9100075725

Just because two function values agree doesn't mean their derivatives have the same value...

That's where your wrong kiddo

if f(x) = g(x)
then f '(x) = g '(x)

is this bait or is Veeky Forums full of brainlets?

>probably more than 1 root

there's exactly 2 roots!!!
Pretty easy to see it desu. The parabola has a minimum at x=0 and has 2 branches pointing upwards so it will cut any function 2 times.

why does people use superscript notation when using exponentials (specially exponentials of matrices, functions or nested exponentials like in your case)?
It triggers my autism. Isn't it much clear to write:

[eqn]x = \frac{1}{\sqrt{3}}\exp\left(\frac{1}{\sqrt{12}}\left(\exp\left(\frac{1}{\sqrt{48}}\cdots\right)\right)\right)[/eqn]

? Or better yet

[eqn]x = \bigcirc_{i=1}^\infty\left[f_i(s)\right] \text{ with } f_i(s) = \frac{\exp(s)}{\sqrt{3\cdot4^{i-1}}}.[/eqn]

You people disgust me.

yes but the contrary doesn't hold true.

Fucking engineers I swear. Get off my board. GET OUT!!!

De que año de carrera eres? O eres underage?

En cualquier caso en primero de física hubo un chico que dijo lo mismo que este retard para resolver una ecuación parecida a la tuya y se quedó con el sanbenito de "el logas" para toda la carrera ya.

Let's say...

f(x) = 2x+1
g(x) = 4x-5

At the point where x is 3, f(x) is indeed equal to g(x). Does that mean their derivatives are the same at x=3?

no

[math]e^x - 3x^2 = 0[/math]
[math]e^x = 3x^2[/math]
[math]e^(x/2) = +/- 3^(1/2) * x[/math]
[math]cos(1/2) + i * sin(1/2) = +/- 3^(1/2) * x[/math]
[math]0.88 + i * 0.48 = +/- 3^(1/2) * x[/math]
[math]0.5 + i * 0.28 = x[/math]

Checking:

[math]^(0.5 + i * 0.28) = 3(0.5 + i * 0.28)^2[/math]
[math]e^(0.5) * e^(i * 0.28) = 3(0.25 + i * 0.28 - 0.08)[/math]
[math]1.65 * (cos(0.28) + i * sin(0.28)) = 0.75 + i * 0.84 - 0.24[/math]
[math]1.65 * (0.96 + i * 0.28) = 0.5 + i * 0.84[/math]
[math]1.58 + i * 0.46 = 0.5 + i * 0.84[/math]

If I didn't round the numbers would work out.

It just do mayne

>e^(x/2) = cos(1/2) + i * sin(1/2)
user...
e^(ix) = cosx + isin(x)
let x = y/(i2)
e^(y/2) = cos(y/(i2)) + isin(y/(i2))

you just have to factor out the 0