- #1

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I have a question where I have to find a two term approximation and i'm at a stage similar to this one and I don't know what I can do.

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- #1

- 73

- 0

I have a question where I have to find a two term approximation and i'm at a stage similar to this one and I don't know what I can do.

- #2

mathman

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What do you mean by breaking it up? There is a power series for ln(1+x)

x-x^{2}/2+x^{3}/3+....

x-x

- #3

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- #4

lurflurf

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We know log(x+1)~x

so expansion about x+1=1->x=0 would be nice

let f(x)=log(x+1)

f(x)~f(0)+f'(0)x+f''(0)x^2/2+f''(0)x^3/6+...

- #5

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Find a two term approximation for the following function near t=1.

h = 5ln(3t+1)

and after substituting u+1 for t

I get:

h = 5ln(3u+4)

Now it gives us the hint: 4 = 3+1

But im stuck after this.

- #6

mathman

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Let v=3u+1, so h=5ln(1+v)~5[v - v

Find a two term approximation for the following function near t=1.

h = 5ln(3t+1)

and after substituting u+1 for t

I get:

h = 5ln(3u+4)

Now it gives us the hint: 4 = 3+1

But im stuck after this.

Take it from there.

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