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MaxT wrote:

Yes - that's what I thought was right - a scalar of 16.5

So, is the operation I wrote out - which I saw someone else do - not a valid operation at all?

I didn't think it was but it left me confused.

It's a perfectly valid operation, just that the answer should have been a scalar (rather than a vector), i.e.

Jack Omar wrote:

In the link about derivatives it is shown that it is 0/0 but I should take a really small difference and then shrink it to 0 so it is the same as 0/0 ???

I don't get that part with shrinking to 0, which is pretty much the same as 0/0.

Hi Jack Omar,

Careful -- taking the limit as something approaches 0 is not quite the same as 0/0! We're only interested in what happens as getsThe question we want to answer is: how do we describe the 'slope' of a function at any given point on its graph? Well, as the pictures you've added show, we can

(a) Take the first point on the curve, find another point on the curve, then draw a straight line through them...

(b) ...then calculate the slope of the line you've just drawn.

Post #1 wrote:

I think I made a mistake in last equation but I don't know why.

Looks OK to me -- isn't the last line the same as the first one?

Post #3 wrote:

No -- the division by is for the previous step. They are saying that:Second question isn't there an error in the result ? There is 2x + Δx, then it is devided by Δx, 2x÷Δx + Δx÷Δx then 2x÷Δx + 1 ? Shrinking it into Δx to 0 I get 2x÷0 + 1 ?

I saw that you asked some questions about integration as well -- but let me know if this makes sense first and then we can move on.

Hi MaxT,

Welcome to the forum.

Almost -- the dot product of two vectors is a scalar, not a vector. In this case we need to add together the results, i.e.

9 + 7.5 = 16.5

Welcome!

Welcome!

Zazu, it's perfectly OK if you'd like to share a link to another maths resource and there is a dedicated subforum -- Maths Teaching Resources -- for doing just that (and we will be happy to support you in doing so).

What isn't OK is for you to pretend to ask for help, return to your post a day later and edit it to insert a hidden link to your website. It gives off the impression that you are using our forum as nothing more than a bulletin board to redirect traffic to your site. Please stop overriding my edits -- if you keep doing so I will have no choice but to restrict your access to the forum.

Just a precautionary note -- while Bob's approach in post #6 is correct, we can only do this because the series converges. In other words, the logic would be:

(1) First prove that the series converges, e.g. by looking at the sequence of partial sums (that's the sum of the first N terms) and showing that that sequence converges as N gets larger.

(2) Now that we know the sum converges, we can then do the rearranging of terms that Bob outlined in post #6 to identify what it converges to.

If the series did not converge then this kind of manipulation is not allowed -- take for example the divergent series 1 - 1 + 1 - 1 + 1 - ... which can be rearranged to 'converge' to lots of different things.

Similar reasoning is also required for differentiating/integrating the infinite series in post #2. We can do this here because p is a probability (so is between 0 and 1, within the series' radius of convergence) and so the series is integrable within that radius of convergence (and its integral also converges).

Hi jadewest,

The key bit is here:

jadewest wrote:

each number increase on the scale indicates an intensity 10 times stronger than the previous number on the scale.

In other words, if you've got an earthquake A measuring 1 on the Richter scale and earthquake B measuring 2 on the Richter scale, earthquake B has an intensity 10 times stronger than earthquake A. Does that make sense?

What would happen if earthquake A measured 2 on the Richter scale and earthquake B measured 4 on the Richter scale -- what would the difference in intensity be then?

More generally any cubic can be reduced to a depressed cubic by substitution -- and then the method above can be applied.

What have you tried? In what context were you given this problem?

The answers to these questions will help us understand what type of solution you are looking for.

Jeremy Desmond wrote:

Thank you Ganesh for that extra interesting information about pi. But you still haven’t explained how we know pi has an infinite number of decimal places containing non-repeating digits. You stated that computers have calculated pi correct to 1.33 x 10^13 decimal places but how do we know the end is not just round the corner? Is it possible that one day a computer will calculate the exact value of pi to a finite number of decimal places?

Hi Jeremy,

Welcome to the forum.

The reason that pi doesn't have a final digit is because it is an irrational number, and irrational numbers have decimal expansions which continue forever. In other words, to answer your question it suffices to:

(1) Prove that pi is irrational, and then

(2) Prove that all irrational numbers have an infinite, non-recurring decimal expansion.

There are quite a few proofs that pi is irrational, some of which are more complex than others. One of the most common proofs is Lambert's, where he essentially (a) shows that all infinite continued fractions are irrational and then (b) finds an infinite continued fraction for pi, which automatically implies that it must be irrational from part (a). If you're interested in continued fractions I've got some videos about it on my YouTube channel (although it doesn't discuss the continued fraction of pi). This takes care of (1).

Now for (2). Suppose instead that you could find an irrational number which didn't have an infinite, non-recurring decimal expansion. Let's say for example, the number 0.12345123451234512345... This decimal expansion has an infinitely recurrent pattern (the '12345' bit). But we see that if we let x = 0.12345123451234512345... then:

Multiplying both sides by 100000:

Subtracting x from both sides:

And finally, dividing both sides by 99999 gives us:

But hang on -- if we can express it as a fraction of two integers, it must be a rational number! (This is pretty much the definition of what it means for a number to be rational.) Similarly, if pi had a 'final digit' we could also express it as a fraction -- for example, if pi terminated after 5 decimal places, i.e. 3.14159, then we could write that as 314159/100000, which is a rational number.

Let me know if that makes sense -- happy to clarify anything if needed.

Abbey78336 wrote:

Give factors of 8.81632653061

8.8132653061 is roughly 432/49, which is the same as the coefficient of the middle term that you wrote in your problem here:

Abbey78336 wrote:

2x^2 + (432/49x) = 3

Exactly what have you been asked to do?

Looks good to me.

You can also type the pi symbol in text form by typing the following:

`:pi`

which produces π.

Hi Bob,

Yes, agreed -- the answer will mirror the binomial expansion of (f + g)^3, with the same binomial coefficients, just with number of derivatives rather than powers.

Hi Bob,

No I don't think you are on the wrong track -- your f1 function works (it is three times differentiable on but not four times) because of the cusp at , and you are right that there are a range of possible solutions we could have just by adding constants (or even polynomial terms that vanish after taking successive derivatives).I think the two questions are intended to be separate (the condition that f, g are three-times differentiable is probably just there to ensure that f''' and g''' exist rather than bearing any relation to Q1). I am interpreting (f.g)^(3) as meaning the third derivative of (f times g) but happy to be corrected by the thread starter.

Hi YHWH,

Welcome to the forum.

YHWH wrote:

1.Find a function f: R → R that is three times differentiable on R but not four times.

It might be easiest to start with a function that isn't differentiable at a single point (say, x = 0) and work backwards.

Consider a function which returns 1 for positive values of x and -1 for negative values. What function could you differentiate to get that?

YHWH wrote:

2.Let the functions f and g: R → R be three times differentiable. Calculate (f .g)^(3).

This is a demonstration of Leibniz's rule for differentiation. Use the product rule to determine the first derivative -- then differentiate again, and again, and see if you can spot a pattern.

Thanks both -- I've edited the original post back to what it was (and it should stay that way).

Hi David,

Is your goal to sit the A-level exams privately or are you looking to self-study the courses for general interest?

In addition, **Disfruta las Matemáticas** is the Spanish version of the website.

Hi jadewest,

Yes, that's right.

One way you can check your answer is just to consider some different fence post heights. Suppose you had a fence post that was 49 inches tall -- this is clearly within the 2 inch margin of error. According to the inequality in option A, we have:

|x - 48| = |49 - 48| = 1

and this is less than or equal to 2, so the inequality works in this case.

Similarly if you choose a fence post height of 47 inches, your inequality will still hold.

If you had a fence post of 51 inches then your inequality shouldn't hold -- and indeed in this case we have |x - 48| = |51 - 48| = 3, which is not less than or equal to 2. So your inequality correctly distinguishes between fence posts within a 2 inch margin of error and those that aren't.

Hi CurlyBracket,

Yes, that's right. In general:

These are sometimes also called the binomial coefficients, because they represent the coefficients (i.e. the 'leading terms') in the expansion of A shorthand way of denoting the above is:which you might see on your scientific calculator.

Note that an important point is whether or not the order matters. For example, suppose you have four cards labelled (A,B,C,D) and you want to choose 2 out of those 4. How many different ways are there to do it? If we don't care about the order, the combinations are just:

(AB, AC, AD, BC, BD, CD)

which represents different ways. On the other hand, if the order does matter then we'd have:(AB, BA, AC, CA, AD, DA, BC, CB, BD, DB, CD, DC)

which represents different permutations. If the order of the objects we've selected is of interest then we use a slightly different formula: