- #1

keebs

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If h(x)~g(x), is h(x+1)~g(x)?

And, if h(x)~g(x), is h(x)h(x+1)~g(x)g(x+1)?

Thanks in advance for any help...

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- Thread starter keebs
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- #1

keebs

- 19

- 0

If h(x)~g(x), is h(x+1)~g(x)?

And, if h(x)~g(x), is h(x)h(x+1)~g(x)g(x+1)?

Thanks in advance for any help...

- #2

Think about [itex]e^{(x+1)}=e \cdot e^x[/itex]; look at the definitions.

- #3

keebs

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Ahhh, ok. Thank you.

Last edited:

- #4

Hurkyl

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If h(x)~g(x), is h(x+1)~g(x)?

I'm not sure that implication holds in general... I can imagine failure can occur if the functions grow sufficiently fast, or if they can do other odd things, like zig-zag back and forth.

- #5

keebs

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I'm not sure that implication holds in general... I can imagine failure can occur if the functions grow sufficiently fast, or if they can do other odd things, like zig-zag back and forth.

What about with the prime counting function? Is pi(x+1)~x/lnx?

- #6

Hurkyl

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- #7

keebs

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Ah, ok. Because if either one of those is true then it implies that pi(x+1)~x/lnx.

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