Solve Numerical Problems: 4x^2 - e^x = 0

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Homework Help Overview

The discussion revolves around solving the equation 4x^2 - e^x = 0 and finding intervals that contain its solutions. Participants also explore the convergence of various sequences as n approaches infinity.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss sketching the graphs of 4x^2 and e^x to identify intersection points, questioning how to determine intervals from these intersections. There is also mention of using the Intermediate Value Theorem (IVT) and evaluating function behavior at infinity.

Discussion Status

The conversation is ongoing with participants sharing hints and suggestions. Some express uncertainty about the methods discussed, while others propose different approaches to identify intervals of intersection and convergence rates.

Contextual Notes

There is a focus on understanding the behavior of the functions involved and the implications of their positivity. Participants are also navigating the complexities of convergence rates for various limits.

stunner5000pt
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Find intervals containing solutions to
[tex]4x^2 - e^x = 0[/tex]
well someone suggested i sketch the graphs of 4x^2 and 2^x but I am not sure on how to go past that point...
All i have to do is find the intervals so do the intersection point(s) of these two functions indicate the intervals where the roots may exist?

Find the Rates of convergence of the following sequences are n -> infinity

a) [tex]\lim_{n\rightarrow\infty} \sin{\frac{1}{n}} = 0[/tex]
i was wondering if expanding the sine by taylor series would help the case.. and if that is so, at which term would i stop at?

b) [tex]\lim_{n\rightarrow\infty} \sin{\frac{1}{n^2}} = 0[/tex]
c) [tex]\lim_{n\rightarrow\infty} (\sin{\frac{1}{n}})^2 = 0[/tex]
d) [tex]\lim_{n\rightarrow\infty} \ln{(n+1)}-\ln{n}= 0[/tex]

the last one looks like ln((n+1)/n) which i think can be expanded by taylor series... if this is the right idea please do tell.
ALso what is the taylor series about.. as in what is the x0 value and the c value where c is the n+1 derivative's evaltuation point.

any help that you could provide for these would be greatly appreicated!
 
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Intersections of the 2 separate functions will BE roots of the given function. Any interval containing an intersection will also contain a root. You can also evaluate the function looking for sign changes, any interval, say (a,b), for which f(a)f(b)<0 holds, contains a root or odd number of roots.
 
im not quite sure how to apply that here...
i mena 4x^2 and e^x are always positive so the f(a)f(b) is never zero

and to pick the intervals of intersection... isn't that just guessing?
 
Hint: add [tex]e^x[/tex] to both sides of the equation.
 
how would adding e^x to both sides help in any way?? What I am trying to do is find the intervals where 4x^2 = e^x..
 
Just to illustrate that what you are looking for is one equal to the other, meaning where the graphs intersect. Or, did I miss something? Looks like you already got that far. So, what do you need, the exact x?
 
Last edited:
well I am trying to find out the intervals and not the points themselves of where the two graphs intersect.
so like integral said i have to pick some interval where these intersections (i think there are two) lie.
 
hmmm

well... Firstly, take the limits of the funtion as they approach -infinity and +infinity. This should show you the behaviour of the graphs as the go "inward". [ Nextly, try dividing the f'n in half to see how it further behaves. (i'm not sure if this will help you much) ]. Furthermore, ever heard of ivt?

[ever try and find solutions to a quintic?]-CALVIN
 
hint: I think there are more than 2, but that's just me.
 
  • #10
im not usre how i can use that ...
yes i have heard of IVT
but is finding the two points just a matter of gueswork? If that were the case then it would be daunting if there were more than two solutions...
 
  • #11
Well, if you do the first part I mentioned, and see how the function behaves outwards, you've done a good part of the work. This leads to IVT, which should be the easy part. It may seem like guesswork (and it is), but who says guesswork is bad? -Calvin
 
  • #12
Oh, and having more than 2 solutions is not "daunting". (Thinking Poly's?)
 
  • #13
ok for the convergence of the limits - is the rate of convergence for the first limit 1/n^2, for the second limit (1/n^4)?
 

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