+0

hi! I want to know how to solve somthing like

0
498
5

hi! I want to know how to solve somthing like

x*x*x*x*x =12

without a calculator

I know that it's equal to x^5 = 12

x = sqrt5(12)

but how can I calculate which number sqrt5(12) is without a calculator?

I really reallly really want to know!!!

/edit

NEWTON'S METHOD! can some1 give me an explanation? or a nice link or something? Would be really helpful!

Guest May 24, 2014

#3
+889
+5

Do you really mean 'without a calculator', or are you simply not allowed to use the power ^ button ? The 5th root of 12 is 1.643752 (6dp). How many decimal are you supposed to calculate ? Whatever numerical method you use there is going to be a fair amount of heavyish arithmetic.

Best of the elementary methods is Newton-Raphson, (the formula for which you will find in the first response to this question).

Using that method,

$$f(x)=x^{5}-12, \text{ so that } f'(x)=5x^{4}$$

and so

$$f(x_{n+1})=x_{n}-\frac{x_{n}^{5}-12}{5x_{n}^{4}}=\frac{1}{5x_{n}^{4}}\left(4x_{n}^{5}+12\right).$$

Starting with x0=1.5, that produces the sequence 1.674, 1.64483, 1.643753, 1.643752, (so convergence to 6dp after 4 iterations).

Good luck if you have to do this on paper !

Bertie  May 24, 2014
Sort:

#1
+676
+5

The Newton method is quite the complicated method if you don't have an advance knowledge of Maths. However, given that you have asked, I will try my best to teach you the Newton Method.

Now then let is get started. The Newton's Method uses a very repetitive method of approaching a root:

$${\sqrt{}}$$

The root that you wish to solve or locate will depend what your initial number will be. In this example I will use the most commonly used pronumeral to replace a number. The equation you are looking for is indicated below:

$$x_n_+_1=(x_n)-f(x_n)/f'(x_n)$$

Looks complicated right?

I'll try to walk you through.

In the equation $$x_n$$ is the value of the $$x$$ value. $$f(x_n)$$ is the same $$x_n$$.

I am very bad at wording... Please excuse me.. I never exceled at Engrish.

$$f'(x_n)$$ is the slope or as known as the derivative, at

$$x_n. x_n_+_1$$

represents the next x-value that you are trying to find.

$$f'(x_n)$$ the derivative represents $$f(x)/dx (dx = delta-x)$$. Therefore, the term $$f(x)/f'(x)$$ represents a value of dx.

Right now.. I am diving into the unknown now... Hopefully, I still know what I am doing >_<

$$\frac{f(x)}{f'(x)}=\frac{f(x)}{f(x)/dx}=dx$$

...and that is all I remember from this method. Anyone could help me out? :P

TakahiroMaeda  May 24, 2014
#2
+5

ThankS :D

It's very easy!

http://youtu.be/PIPiv6gn_Ls

my school teaches us an "ABC-formule"

which I need to solve things like x^3 +2x +5 =0 with

THIS METHOD IS MUCH BETTER IMO :D

it basicly estimates roots :D!!!

veeeryy handy

But I see that it can also give wrong estimates sometimes

:D

Guest May 24, 2014
#3
+889
+5

Do you really mean 'without a calculator', or are you simply not allowed to use the power ^ button ? The 5th root of 12 is 1.643752 (6dp). How many decimal are you supposed to calculate ? Whatever numerical method you use there is going to be a fair amount of heavyish arithmetic.

Best of the elementary methods is Newton-Raphson, (the formula for which you will find in the first response to this question).

Using that method,

$$f(x)=x^{5}-12, \text{ so that } f'(x)=5x^{4}$$

and so

$$f(x_{n+1})=x_{n}-\frac{x_{n}^{5}-12}{5x_{n}^{4}}=\frac{1}{5x_{n}^{4}}\left(4x_{n}^{5}+12\right).$$

Starting with x0=1.5, that produces the sequence 1.674, 1.64483, 1.643753, 1.643752, (so convergence to 6dp after 4 iterations).

Good luck if you have to do this on paper !

Bertie  May 24, 2014
#4
+91435
0

---------------------------------------------------

"But I see that it can also give wrong estimates sometimes

---------------------------------------------------

Melody  May 24, 2014
#5
+889
0

We are solving the equation f(x)=0. Graphically the roots (solutions) of the equation are the x co-ordinates of the points where the y=f(x) crosses the x axis, i.e. the points where y = 0.

The Newton-Raphson method starts with a point x0 close to a root, takes the tangent to the curve for that value of x, (the tangent at A in the sketch below) and then calculates the x co-ordinate of its intersection with the x axis, (x1 in the sketch below).  (All done algebraically of course). Hopefully x1 will be a better approximation to the root  than was x0. The process is then repeated using x1 as the new starting point to find an x2 then again to find an x3 and so on until converged to whatever degree of accuracy is required.

The method can go awry if the gradient of the tangent is small, it's possible to be kicked a long way (possibly irretrievably) from the original point. In the sketch below, for example, suppose x0 were further to the right so that the tangent to the curve was constructed at B. The next approximation could be way off to the left or the right depending on whether B is to the left or right of the maximima. Either way the next 'approximation' to the root will be useless. On paper, the problem manifests itself with a small value for f '(x) and a correspondingly large correction term f(x)/f '(x).

Bertie  May 24, 2014

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