Need help with a difficult problem
Given that x is so small that x^3 and higher powers of x can be neglected, show that 1/sq. root of (1+x) = 1 - x/2 + 3x^2/8
By letting x= 1/4, find the rational approximation of sq. rt. of 5.
Does anyone know how to solve this? Thanks in advance. The ^ symbol indicates the degree. I'm sorry if this problem is not game related but I figured there are a lot of math experts here. I hope someone can help.
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[edited by - majorbino on February 3, 2003 6:11:51 AM]
Got Blog? Visit the BlogBoard at http:users.boardnation.com/~ace
That isn''t equal to, it is approximately equal to. That is the first three terms of the second order taylor series for that equation. Plug 1/4 into 1/(1+x) and what do you get? Now exactly what have you done to try to work this homework problem yourself?
Keys to success: Ability, ambition and opportunity.
This sounds very much like a homework problem, which means telling you the answer is not helping you, since you will not learn anything from that. Besides, it always feels much better to solve a difficult problem yourself, than having someone tell you how to.
However, I can give you a hint. Look at what you''re given: 1/sqrt(1+x) = 1 - x/2 + (3x^2)/8 obviously must be true if you ignore polynomial terms with exponent three or more, since you are asked to prove it. So, if you keep those terms, it must look something like 1/sqrt(1+x) = 1 - x/2 + (3x^2)/8 + ax^3 + bx^4 + cx^5 + ... (at least, we have not lost any generality by assuming that). I bet you''ve seen something similar to this in your math text book, if you have any.
However, I can give you a hint. Look at what you''re given: 1/sqrt(1+x) = 1 - x/2 + (3x^2)/8 obviously must be true if you ignore polynomial terms with exponent three or more, since you are asked to prove it. So, if you keep those terms, it must look something like 1/sqrt(1+x) = 1 - x/2 + (3x^2)/8 + ax^3 + bx^4 + cx^5 + ... (at least, we have not lost any generality by assuming that). I bet you''ve seen something similar to this in your math text book, if you have any.
I would''ve answered the problem myself but I really got stuck with this one. There''s nothing and no one to help me.
Since you know the answers, could you at least tell me how to work on a problem like this?
By the way, this Math board rocks. I''ve been trying to find one of these for a long time.
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By the way, this Math board rocks. I''ve been trying to find one of these for a long time.
Got Blog? Visit the BlogBoard at http:users.boardnation.com/~ace
February 03, 2003 06:06 AM
I''ll make two suggestions...
1. Work on paper using the real sqrt root sign. I''m sure you are but it''s much easier that way. The best waay to do maths is to not just look at the problem but play around with it.
2. Now, second and more useful tip, you''ve got sqrt(1+x) and you know the value of x to put in x = 1/4. You also know the value you want inside the sqrt root (ie you want to see sqrt(5) in that equation... can you play around to convert sqrt(1 + 1/4) into something like sqrt(5).
You''ll kick yourself if we tell you, because it''s really easy. Honest!
1. Work on paper using the real sqrt root sign. I''m sure you are but it''s much easier that way. The best waay to do maths is to not just look at the problem but play around with it.
2. Now, second and more useful tip, you''ve got sqrt(1+x) and you know the value of x to put in x = 1/4. You also know the value you want inside the sqrt root (ie you want to see sqrt(5) in that equation... can you play around to convert sqrt(1 + 1/4) into something like sqrt(5).
You''ll kick yourself if we tell you, because it''s really easy. Honest!
Thanks a lot everyone. I know it now. The answer is 256/115. 
There''s a 0.01 difference though when I tried sq. rt. of 5 in my calculator. That''s negligible, isn''t it?
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There''s a 0.01 difference though when I tried sq. rt. of 5 in my calculator. That''s negligible, isn''t it?
Got Blog? Visit the BlogBoard at http:users.boardnation.com/~ace
majorbino,
I''m very happy that aleph and Anonymous Poster gave you advice on how to solve the problem, rather than the answer. That was the right way for them to answer your question!
I appreciate all the folks in this forum who help folks who help themselves, and it sounds like their advice helped you to find the answer for yourself!
FYI, you can review the forum policy on homework here:
Forum FAQ
Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.
I''m very happy that aleph and Anonymous Poster gave you advice on how to solve the problem, rather than the answer. That was the right way for them to answer your question!
I appreciate all the folks in this forum who help folks who help themselves, and it sounds like their advice helped you to find the answer for yourself!
FYI, you can review the forum policy on homework here:
Forum FAQ
Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.
Graham Rhodes Moderator, Math & Physics forum @ gamedev.net
Well, I gave a hint too
It was obtuse though. I''ll try to make is slightly clearer. It is approximately equal so the differance is appoximately zero. You have to prove that the differance is less than x^3. That is your given. Anything within x^3 of the right answer is good enough. So if you can prove the differance is less than x^3 you proved it. I''ll give you one more hint, get rid of the square root. Unless of course "I know it now" means you proved the first part in which case nevermind.

Keys to success: Ability, ambition and opportunity.
quote:
Original post by LilBudyWizer
Well, I gave a hint tooIt was obtuse though.
Dude, I''m sorry I didn''t include your name. Didn''t actually see your post, since it was shorter than the others and stuffed in the middle. Yes, you did indeed give advice. And hey, I always appreciate your contributions to this forum,
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Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.
Graham Rhodes Moderator, Math & Physics forum @ gamedev.net
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