Mathematics Corner :)

The Ruler said:
Iteration is relatively easy.
.
.
.
.
.
.
Until the results that you get don't show the change in sign. That drives me mad because I (still) can't figure out what I'm doing wrong.
Click to expand...

Do you always substitute the two values back into the original function?
 
Solving using Iterations is not always easy, especially when a function is not well behaved.
Also choosing the initial point is very important or the solution might not converge
 
Oh, and make sure the equation that you're putting values into equals zero (rearranging whatever original expression you're given to make sure that's the case...)??

Osman said:
I, on the other hand, actually loved it. :-[
Click to expand...

Lol, really?! I dread anything even vaguely linked to using trial & improvement...
 
GreyKode said:
What change in sign? That's not always the case.
Click to expand...

Sign-change doesn't technically come under iteration, lol...but it's a simple method we're taught for approximating solutions...if you're trying to solve x^5 = 5x, for example, you rearrange to get x^5 - 5x = 0, then try values for x which give answers of roughly 0...substituting in x = 1 gives you -4; using x = 2 gives you 22...by the sign-change rule, you know x lies between 1 and 2, lol.

In iteration, obviously, you rearrange the equation to get x=..., and repeatedly put values in...choosing an initial value that'll make the iteration converge becomes much more important, and annoying, there...
 
Last edited:
Question:

DSCF2526.jpg


My attempt at answering it:

DSCF2525.jpg


... And I'm stuck.

EDIT: I just realised my (1+x^2) had turned to (1-x^2) suddenly. *sigh*
 
Last edited:
LOL got your solution right here

you used the wrong substitution

let x =tan(c)
c goes from 0 ---> tan^-1(1/4)
1+x^2=sec^2(c) dx=sec^2(c)dc

The inegrand becomes [1/sec^3(c)] *(sec^2(c)) = cos(c)---> integrate(cos(c))
 
Last edited:
No it's a given substition. You can't use any other substitution. x=sin theta was given.
 
consider sin^2 = sin^2(theta), sin = sin(theta) so it's not too cluttered:

integral dx/(1+sin^2)^3/2--> integral dx/(sqrt(1+sin^2))^3

(remember 3/2 = 3*1/2, x^1/2 = sqrt(x), sqrt = square root)

--> integral dx/(1+sin)^3

and so on and so forth, I think atleast,.
 
Last edited:
You're resorting to all sorts of media, lol...I like your handwriting...
Are you suuure it's not in fact (1 - x^2)^3/2??...because otherwise, I agree with GreyKode; the substitution used when the bottom is (1 + x^2) usually involves tan...

If it IS (1 - x^2), the only thing wrong with your working is where you've written 1/cos u dx ...(not to add the the confusion, but I'm gonna use u instead of theta :X).....because dx/du = cos(u), you replace the dx in the integral with cos(u) du....ultimately, you want everything in the integral to be in terms of u...
So, the integral will just be:

1/(1-sin^2(u))^3/2 * cos(u) du

= 1/cos*3(u) * cos(u) du...........(basically, the last cos(u) is at the top, not the bottom)

= 1/cos^2(u) du = sec^2(u) du >> integrated, this gives you tan(u)

Also, the new limits are pi/6 and 0, I think; evaluate tan(u) between these... all this assuming the question may be wrong, lol... :blind:
 
ok using alcurad's notation

starting from

integral(sec^4)

sec^4=(sec^2)(sec^2) = (1+tan^2)(sec^2)

but sec^2d_theta= dtan

therefore

the integral becomes integtral(1+tan^2)dtan

now its simple let tan(theta) = y and change limits
 
That is how I understood it, but I was hoping it wasn't just wishful thinking that made you change it to a 1-x^2 in your working, lol...if it definitely isn't a minus, and you have to use that substitution, I'm not really sure...

Do you by any chance have the answers? It's not (sqrt3)/3??
 
~Raynn~ said:
= 1/cos*3(u) * cos(u) du...........(basically, the last cos(u) is at the top, not the bottom)
Click to expand...

This is where I went wrong.

But when would you use dx/du instead of du/dx?

I don't have the answer... But I think you're right. Because I re-did it without looking at the answer and (sqrt3)/3 is what I got too.
 
I think you went wrong all the way from the beginning
where you changed 1+x^2 to 1-x^2.

about the du/dx stuff

Here's the rule

integral f(x) dx----(1),
if you make a sub x = g(u), then you should replace every x in (1) with u
therefore
f(x)--->f(g(u))
since x=g(u)

then dx/du = d/du g(u), multiplying both sides by du, you get dx = d/du g(u) du---(2)

so now the integral becomes integral f(g(u)) d/du g(u) du
 
GreyKode said:
I think you went wrong all the way from the beginning
where you changed 1+x^2 to 1-x^2.
Click to expand...

The question was originally (1-x^2) :exhausted

about the du/dx stuff

Here's the rule

integral f(x) dx----(1),
if you make a sub x = g(u), then you should replace every x in (1) with u
therefore
f(x)--->f(g(u))
since x=g(u)

then dx/du = d/du g(u), multiplying both sides by du, you get dx = d/du g(u) du---(2)

so now the integral becomes integral f(g(u)) d/du g(u) du
Click to expand...

Aha. Thanks.
 

Similar Threads

Back
Top