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Quick calc problem help plz.

D

DoNotDisturb

Senior member
Ok, theres no example in the book and i'm kind of stuck:

the problem is:

g(x) = integral from 1 to x is (1/t)dt

Conjecture: g(x) = ln x

test this conjecture by finding g'(3) numerically. How does your result confirm the conjecture.

________________________________________________________________________________

ok... how am i supposed to find the derivative i believe, because it has the prime symbol of this exponential function.
 
I'm in calc II in college....and I am lost. Maybe its because it is written out in text instead of symbolicaly...but that makes no sense to me.
 
ive taken calc 1,2,3, theroitical physics, electrodynamics & plasmas, Diff EQ's and Quantum Mechanics. and ive never heard teh term "Conjecture". WTF is that

teh integral of (1/t) from 1-x is Ln(x). thats simple. dont have a clue what teh rest is even asking
 
Originally posted by: DoNotDisturb
Ok, theres no example in the book and i'm kind of stuck:

the problem is:

g(x) = integral from 1 to x is (1/t)dt

Conjecture: g(x) = ln x

test this conjecture by finding g'(3) numerically. How does your result confirm the conjecture.

________________________________________________________________________________

ok... how am i supposed to find the derivative i believe, because it has the prime symbol of this exponential function.
Click to expand...


Your book has a weird way of writing things.

What I THINK it's trying to say :

The derivative of ln x = 1/x.
We conjecture that the integral of (1/t) from 1->x = ln x.

This is true. The integral of (1/t)dt = ln t. Using the bounds as 1 and x gives you ln x - 0.

To get it numerically, take the integral of (1/t) from 1->3.

Done.

[EDIT] From 1->3 . Sorry, 0 gives you negative infinity 🙂 [/EDIT]
-silver

 
Originally posted by: TheEvil1
ive taken calc 1,2,3, theroitical physics, electrodynamics & plasmas, Diff EQ's and Quantum Mechanics. and ive never heard teh term "Conjecture". WTF is that

teh integral of (1/t) from 1-x is Ln(x). thats simple. dont have a clue what teh rest is even asking
Click to expand...

conjecture is an estimate that is made.. for example if you have 2 values, and you're trying to find the exact area of lets say, a function.

evaluating 1 and 3 of this gives you 1 and 3 for the area (just using this as an example)... they tell u what u think x =2 's area will be.... well since the area is equal to x, then u can conjecture, or 'estimate' that at x=2, the area will be 2.
 
Originally posted by: DoNotDisturb
Originally posted by: TheEvil1
ive taken calc 1,2,3, theroitical physics, electrodynamics & plasmas, Diff EQ's and Quantum Mechanics. and ive never heard teh term "Conjecture". WTF is that

teh integral of (1/t) from 1-x is Ln(x). thats simple. dont have a clue what teh rest is even asking
Click to expand...

conjecture is an estimate that is made.. for example if you have 2 values, and you're trying to find the exact area of lets say, a function.

evaluating 1 and 3 of this gives you 1 and 3 for the area (just using this as an example)... they tell u what u think x =2 's area will be.... well since the area is equal to x, then u can conjecture, or 'estimate' that at x=2, the area will be 2.
Click to expand...

ok i understand that but its still liek greek. and i also think its funny that ive never used it or herd it before
 
Originally posted by: agnitrate
Originally posted by: DoNotDisturb
Ok, theres no example in the book and i'm kind of stuck:

the problem is:

g(x) = integral from 1 to x is (1/t)dt

Conjecture: g(x) = ln x

test this conjecture by finding g'(3) numerically. How does your result confirm the conjecture.

________________________________________________________________________________

ok... how am i supposed to find the derivative i believe, because it has the prime symbol of this exponential function.
Click to expand...


Your book has a weird way of writing things.

What I THINK it's trying to say :

The derivative of ln x = 1/x.
We conjecture that the integral of (1/t) from 1->x = ln x.

This is true. The integral of (1/t)dt = ln t. Using the bounds as 1 and x gives you ln x - 0.

To get it numerically, take the integral of (1/t) from 1->3.

Done.

[EDIT] From 1->3 . Sorry, 0 gives you negative infinity 🙂 [/EDIT]
-silver
Click to expand...

what do i do then?
 
Originally posted by: DoNotDisturb
Originally posted by: agnitrate
Originally posted by: DoNotDisturb
Ok, theres no example in the book and i'm kind of stuck:

the problem is:

g(x) = integral from 1 to x is (1/t)dt

Conjecture: g(x) = ln x

test this conjecture by finding g'(3) numerically. How does your result confirm the conjecture.

________________________________________________________________________________

ok... how am i supposed to find the derivative i believe, because it has the prime symbol of this exponential function.
Click to expand...


Your book has a weird way of writing things.

What I THINK it's trying to say :

The derivative of ln x = 1/x.
We conjecture that the integral of (1/t) from 1->x = ln x.

This is true. The integral of (1/t)dt = ln t. Using the bounds as 1 and x gives you ln x - 0.

To get it numerically, take the integral of (1/t) from 1->3.

Done.

[EDIT] From 1->3 . Sorry, 0 gives you negative infinity 🙂 [/EDIT]
-silver
Click to expand...

what do i do then?
Click to expand...

I already told you how to do it.

They're saying they THINK that the integral of (1/t) from 1->x == ln x.

You're trying to prove that.

So since you know the integral of (1/t) IS ln x, you already know that solving that integral gives you ln x - ln 1 = ln x - 0 = ln x.

There's your proof.

Now I'm not sure what the actual problem they want you to do is. If it is g`(x), that means the first derivative of g. Since g = integral (1/t) from 1->x, taking the derivative of an integral just kinda cancels them out so you get 1/t or 1/3 in your case.

If that ISN'T g prime of x, then I dunno what to tell you. I'm 99% sure it is though since you're only in AB.

Have fun.

-silver
 
in the back they solved it like this:

g'(3) = g(4) - g(2) /2 = 1.38629 - 0.69314 / 2 = 0.3466 apprx equal to (d/dx)(ln x) which is 1/3

how did they come up with those numbers.
 
You have to use the approximation of ln(x) and derive that. Then insert 3 where the x's are and solve that.
 
Definition of slope:
g'(x) ~= [g(a+b) - g(a)]/b as b approaches zero

I think this is what they wanted you to use when they said to solve it numerically.

so you have:
g(x) = ln x

g'(3) ~= [ln(2 + 2) - ln(2)] / 2 = [ln(4) - ln(2)] / 2 ~= 0.3466

maybe a better approximation would be to make b smaller like 0.1

g'(3) ~= [ln(2.95 + 0.1) - ln(2.95)] / 0.1 = [ln(3.05) - ln(2.95)] / 0.1 ~= 0.3334



next you have:
g(x) = integral from 1 to x of (1/t)dt

g'(x) = (d/dx) g(x) = (d/dx) integral from 1 to x of (1/t) dt = 1/x
g'(3) = 1/3 = 0.333333... repeating

[Edit: fixed typos]
 
thats the thing, we HAVEN'T done, natural logs yet, haven't dealt with ln x or ANYTHING. Thus, you can't use ln x to solve for ANYTHING, because we're assuming that you do not know that the derivative of it is 1/x..... they used the symmetric diff quotient to conjecture that g(x) = ln x

how do we know?

the equation is

nDer is g'(x) = g(x+h) - g(x-h) / 2h

solving for g'(3), , since I already have the values in whole numbers, i'll make the tolerance = 1

so that yields to: g(3+1) - g(3-1) / 2*1 = g(4) - g(2) / 2

my prev post, discusses the value of this. That is how you solve for the numerical derivative algebraically, basically you can do a backward, forward, or symmetric diff quotient to achieve this, however, symmetric is much more accurate than the forward or backward.

edited: denominator of symmetric diff quotient
 
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