Polynomials - Halving the Interval/Bisection (1 Viewer)

[rampeh]

When they say find the root of a polynomial given some range, to two decimal places , How do you choose when to cut your losses and round up/down?

Is it when the solution of the root is 0.00*, or as soon as the divided root becomes two decimal places long? =|

Also, how do you round it.

Thanks in advance

 

[pLuvia]

Never round up until the final answer, i.e. for each answer you get before hand don't round it use the full decimal answer, it's until the end where you get the final answer you round the answer to the required decimal place.
Even when the number is 76, if it says two decimal places you must write it as 76.00

 

[Riviet]

Also, use your calculator's memory key/function to save those exact values.

If your final answer is something like:

0.2593392482...

and the questions asks for rounding the answer to 2 decimal places, then you would count two decimal places to the right of where the decimal point is so your answer would be 0.26 since the 4th digit from the left, 9, is greater than or equal to 5 so you have to round up for the previous (3rd in this case) digit (changing the 5 to a 6).

Hope that helps.

 

[tristambrown]

with all the ones ive done, when two places are specified, i keep going until the i get to two decimal places that cease to change and then round those..

ie the halves (that i have pulled out of my head from nowhere) proceed like follows

2.36569978
2.3256798
2.3536799998
2.339676868609
2.34750549409
2.3484844059

then round 2.348 or 2.347 to 2.35 (the fact that both answers will round to the same thing when rounded to the two places means that you are now accurate enough and can now stop the halving process) - anything before this (unless specified) will end up being too innacurate... ..

personally i prefer newtons method - far quicker, more accurate and less steps - use that if you have the option

 

[Riviet]

personally i prefer newtons method - far quicker, more accurate and less steps - use that if you have the option

In most HSC questions I've seen in past papers, the question usually will ask you to use a specific method. Make sure you follow exactly what the questions asks for. If it doesn't specify which method, then I guess you're free to use either, Newton's method being the better one for speed and accuracy reasons.

 

[STx]

btw, for Newton's Method, do you guys find it easier to find a general formula even when they dont ask and to use it instead of manually entering the data in the calculator everytime?

 

[sasquatch]

I'll tell you a really good way of performing newtons method on the calculator. Cambridge shows that you can find a formula as STx was saying to make it easier to find the root, but instead you can do something like this.

say f(x) = x² - 2x + 1, and a root lies near x= 0.9
f'(x) = 2x - 2

so by the formula for newtons approximation

x₁ = x₁ - f(x) / f'(x).

This is how i would perform it on the calculator.

1. Enter [0][.][9] on the calculator and press [=].
2. Then type the following:
[ANS][-][(][(][ANS][²][-][2][ANS][+][1][)][/][(][2][ANS][-][2][)][)]

to give: Ans - ((Ans² -2Ans + 1)/(2Ans - 2))

3. You can then repeatedly press [=] to give the most accurate approximation for the root.. and if you test it out it gets really close and in several of instances exact.

For this you end up getting 0.999994021, so taking it to 4 decimal places you get 1.

 

[who_loves_maths]

3. You can then repeatedly press [=] to give the most accurate approximation for the root.. and if you test it out it gets really close and in several of instances exact.

For this you end up getting 0.999994021, so taking it to 4 decimal places you get 1.

1) Newton's formula is a contraction mapping on the metric space R¹ with the standard metric for appropriate functions and carefully chosen initial value inputs.

- It will never give you an exact root, unless your intial input is the root.

2) Taking 0.999994021 to be 1 as a root (or even worse, an 'exact' root) is based on the assumption that the fixed point of the contraction mapping is 1 - i.e. that the infinite sequence of continual approximations converges to 1.

In most instances (e.g. exams), these assumptions are fine. But in the case where the limit of the approximation sequence is some transcendental numbers very close to 1, or some other rationals, then clearly taking x = 1 as a root is incorrect (it's becomes worse when the actual root is so close to 1 that even a calculator has to round it to 1).

But these far-fetched situations obviously will never happen in exams, so for those who only care about passing exams, continue doing what you're doing.

 

[sasquatch]

I dont think i made it clear with what i meant by exact. When i said exact it turns out like this due to the rounding of the calculator.

I dont understand your point where the solution becomes mildy incorrect when the calculator rounds the point off to a whole number (even if the root isnt truely a whole number). If the question says find an approximation of the root to 2 d.p. how is saying an approximate value of the root is (1.00) incorrect in comparision to say (0.99). I wasnt guiding people to say the root is "exactly" 1, i was saying that by using this repeated process of finding the root, the "closest" approximation to be possibly achieved using the calculator.

Also i guess i didnt state something else. I wasnt suggesting also to ALWAYS repeatedly press the equals sign until the answer stops chaning, but say the question asks to find an approximation by using newtwons method say 5 times.

Instead of retyping in the formula into the calculator or bothering to symplify the approximation expression, you may press the equals sign 5 times, minimising the time you spend on such a question and hence maximising time efficiency.

Sorry if my original statement was misunderstood/misinterperated by anybody. I didnt mean to give useless/incorect guidance.

 

[who_loves_maths]

^ No sorry sasquatch, I understand what you were trying to do and my fault for not distinguishing that from what I said.

It wasn't my intention to say that you were wrong or being intentionally misguiding lol
I was just trying to make a point about Newton's Method, I probably shouldn't have quoted you like that haha .

 

[sasquatch]

um hehe no problem..??