Manipulating powers - Page 2

ATH · 2008-07-14, 10:02

If y=k*xⁿ and you change the k-value from say 2 to 3, then y will be 1.5 times bigger for all x-values.

But if you change n from say 2 to 3 you get y=k*x*x*x instead of y=k*x*x, so y will be x times higher for all x. For x=1000 y=k*x³ will be 1000 times bigger than y=k*x², and for x=1000000 y=k*x³ will get 1000000 times bigger than y=k*x². No change in k-value will be able to correct a change in n-value.

In general for a>1 any power function y=x^a will get bigger than any linear function y=a*x when x -> infinity. Even if you choose a huge constant in the linear function and a tiny constant in the power function like: y=100000000000*x and y=x^{1.00000000001} the power function eventually get bigger when x->infinity.

Similarly for a>1 any exponential functions y=a^x will outgrow any power function y=x^a as x -> infinity.

retina · 2008-07-14, 10:13

Quote:

Originally Posted by ATH

... and y=x^{0.00000000001}

Perhaps you mean y=x^{[b]1[/b].00000000001}?

ATH · 2008-07-14, 10:23

yes :) sorry

R.D. Silverman · 2008-07-14, 11:26

Quote:

Originally Posted by roger

I didn't want to turn it into a straight line, I just wanted to change the exponent to a whole number, and I thought that I could solve for the k-value with n=2 so that the curve would be the same.

Oh well, guess not.

I am curious why you believed that you could.

Consider a straight line. (exponent = 1). How could anyone with two
working brain cells believe that you could change the exponent to something
other than 1 and have the result still be a straight line??

davieddy · 2008-07-14, 14:37

Quote:

Originally Posted by R.D. Silverman

I am curious why you believed that you could.

Consider a straight line. (exponent = 1). How could anyone with two
working brain cells believe that you could change the exponent to something
other than 1 and have the result still be a straight line??

y = k*x^0

roger · 2008-07-14, 17:23

Thanks for your replies and good advice. I was not trying to drastically change the power, only by about 5% or so, so I guess I thought I could compensate with the coefficient.

This is the curve I was trying to express as a different power, for the twin prime project going on with gdbarnes and co.

davieddy · 2008-07-14, 18:13

Roger, How old are you?
There may be a bit of "learning to run before you can walk"
going on here.

David

roger · 2008-07-14, 18:41

Yeah, I know what you mean. Unfortunately, we don't learn the more interesting stuff until grade 12 and university. I just finished 11. But school is so slow, so I'm trying to learn stuff over the forums and so on. Do you know of some good resources (especially internet)?

Thanks!

R.D. Silverman · 2008-07-14, 18:45

Quote:

Originally Posted by roger

Thanks for your replies and good advice. I was not trying to drastically change the power, only by about 5% or so, so I guess I thought I could compensate with the coefficient.

This is the curve I was trying to express as a different power, for the twin prime project going on with gdbarnes and co.

You lack a basic understanding of even 1st year junior high school
algebra, yet you are working on twin primes and trying to do curve
fitting??????

You are trying to run before you can walk.

ewmayer · 2008-07-14, 21:42

I think the answer is "take logarithms", but I'm not sure what the question is. :)

roger · 2008-07-14, 23:06

Quote:

I think the answer is "take logarithms", but I'm not sure what the question is. :)

Thought so, thanks :)

And @Silverman, you're not dissuading me with your arrogance BTW

2008-07-14, 10:02	#12
ATH Einyen Dec 2003 Denmark 2×1,579 Posts	If y=kxⁿ and you change the k-value from say 2 to 3, then y will be 1.5 times bigger for all* x-values. But if you change n from say 2 to 3 you get y=kxxx instead of y=kxx, so y will be x times higher for all x. For x=1000 y=kx³ will be 1000 times bigger than y=kx², and for x=1000000 y=kx³ will get 1000000 times bigger than y=kx². No change in k-value will be able to correct a change in n-value. In general for a>1 any power function y=x^a will get bigger than any linear function y=ax when x -> infinity. Even if you choose a huge constant in the linear function and a tiny constant in the power function like: y=100000000000x and y=x^{1.00000000001} the power function eventually get bigger when x->infinity. Similarly for a>1 any exponential functions y=a^x will outgrow any power function y=x^a as x -> infinity. Last fiddled with by ATH on 2008-07-14 at 10:23*

2008-07-14, 17:23	#17
roger Oct 2006 260₁₀ Posts	Thanks for your replies and good advice. I was not trying to drastically change the power, only by about 5% or so, so I guess I thought I could compensate with the coefficient. This is the curve I was trying to express as a different power, for the twin prime project going on with gdbarnes and co. Attached Thumbnails

2008-07-14, 18:13	#18
davieddy "Lucan" Dec 2006 England 2×3×13×83 Posts	Roger, How old are you? There may be a bit of "learning to run before you can walk" going on here. David Last fiddled with by davieddy on 2008-07-14 at 18:15

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2008-07-14, 10:23	#14
ATH Einyen Dec 2003 Denmark 2×1,579 Posts	yes :) sorry

2008-07-14, 18:41	#19
roger Oct 2006 2²·5·13 Posts	Yeah, I know what you mean. Unfortunately, we don't learn the more interesting stuff until grade 12 and university. I just finished 11. But school is so slow, so I'm trying to learn stuff over the forums and so on. Do you know of some good resources (especially internet)? Thanks!

2008-07-14, 21:42	#21
ewmayer ∂²ω=0 Sep 2002 República de California 19·613 Posts	I think the answer is "take logarithms", but I'm not sure what the question is. :)