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Hi Pema,

What do you get when you add A and B together?

What kind of matrix is it, and what do we know about powers of these matrices?

Welcome, sarahparker!

and a sequence

such that is the nth prime. It can be shown that this recurrence relation generates all the prime numbers -- however, the complexity of this problem (as is often the case with prime-generating constants) is determining the value of to a sufficiently high degree of accuracy. The proof of this result uses Bertrand's postulate.Here, I've used to denote the floor of x (you can think of that as 'rounding down x to the nearest whole number') and to denote the fractional part of x. So in other words, we'd have and .The exact value of can be represented as an infinite sum:You need about 25 terms in the series above to get all the primes less than 100, for example.

Welcome!

Hi Zeeshan 01,

Is there something missing from your post?

Seems to be a duplicate of this post: http://www.mathisfunforum.com/viewtopic.php?id=24491

Welcome!

Hi Hannibal lecter,

That's the union symbol for sets. The union of two sets is another set containing each of the elements of the individual sets. For example:

In this case, the union symbol means this:

Hi all,

This thread is an exact copy of **this thread**. (As you can see, the poster above simply copied the plaintext from that thread, including the signature. Some of the other links the post contained were also removed, probably because the forum doesn't let you post too many links if you're a new user.) Having googled the topic starter's username, they appear to be a run-of-the-mill bot, registering on multiple forums within the last couple of days and creating one single post, taken from somewhere else on the forum it posted on. This is a common theme for spammers or bots who want to spread their links inconspicuously.

For example, the poster most likely used *that* post because it was created five years ago, at a time which most of us probably wouldn't remember. It therefore becomes a suitable candidate for hiding a link somewhere in, which can attract spammers, or even worse, if accidentally clicked, could cause harm to an unsuspecting victim. No hyperlinks were disguised in this post, thankfully. Sometimes bots also make multiple posts like this before their account gets link-posting privileges, at which point, they can post as much spam as they like until someone stops them. Both of these things happen surprisingly often, and can be quite messy to clean up! Spammers will try all sorts of tricks to get their posts out! One annoying fellow earlier today decided to mix Arabic letters in with the English letters in their username so I could not highlight it as usual in order to ban them.

I won't delete the post or ban them yet (on the slight off-chance they might not have bad intentions), but will keep an eye on them. I think it serves as a useful demonstration of the kinds of threats that our forum can potentially be vulnerable to.

Thanks MathsIsFun, all is good now.

**zetafunc**- Replies: 5

The [ url ] tags no longer function properly: instead of turning a word into a link, it makes the word disappear completely. For example, see this post: http://www.mathisfunforum.com/viewtopic.php?id=24999

I have entered this:

```
Hi,
You might find this link useful: [url=https://www.mathsisfun.com/algebra/polynomials-multiplying.html]https://www.mathsisfun.com/algebra/polynomials-multiplying.html[/url]
```

But the output has been this:

```
Hi,
You might find this link useful:
```

Can we fix this? Any links appearing in any other post have also retroactively disappeared (take a look at this thread, for example: http://www.mathisfunforum.com/viewtopic.php?id=22506 ).

For part (a), what happens when you multiply that matrix by each of those vectors?

For part (b), you'll have been able to find some number (called a 'scalar multiple') such that , and likewise for the others. This then tells you that . Use this to write down an expression for and therefore .For part (c), trial and error is easiest here: try to find a way of making 10 from a linear combination of and . Then see if that same linear combination works for the other two vector components.For part (d), you can use your linear combination found in part (c) to replace that vector with something that looks like . You can then split the matrix multiplication into three parts: can you see how?Welcome, Xavier!

Hi Lasson,

Can you please check your other percentages thread?

Hi,

You might find this link useful: https://www.mathsisfun.com/algebra/poly … lying.html

Hi Alexandr2,

I responded to your other post -- please have a look when you have time.

One of them secured 9 marks more than the other

Let me know if you get stuck.

Hi Russia,

Have you considered registering an account with us? You can do this by **clicking here**.

The old price of the coat is $20, and the new price is $15. That means the price has reduced by $5 (since 20 - 15 = 5). So now you just need to find out how much percent $5 represents as a fraction of the old price. This is:

Please let us know if you need any further help!

you have this instead:

Another way of writing this is:

Now, let's say you wanted to solve this system for and . You'd need to find the inverse of , right? But for that inverse to exist, the determinant can't be equal to 0. In other words, for that thing to have a solution, you must have:i.e.

which tells you that isn't a multiple of . Remember, saying that and are scalar multiples of each other just means that you can find some real number such that , or in other words:i.e.

Now, for the forwards direction (any vector can be expressed as => and aren't multiples of each other), try proving the contrapositive, i.e. show that if for some real number , then not every vector can be expressed in the form . Let me know if you have any more questions!No worries -- let's look at part (a) first.

!nval!d_us3rnam3 wrote:

Okay, let's take any two-dimensional vector, say, . We want to know: can we find real numbers and so that:(a) Show that any two-dimensional vector can be expressed in the form

where and are real numbers.

If we 'multiply out' the left-hand side, we get:

Now, we can add two vectors just by adding the matching components, so that:

In other words, we want to find real numbers and so that:which is exactly the same as solving the pair of simultaneous equations:

Remember, we're solving for and here. (Just pretend that and are any old real numbers.)Let me know if this makes sense -- happy to explain anything further if you need more help.

Hi !nval!d_us3rnam3,

Thanks for your post -- I fixed your LaTeX.

For part (a), suppose you've got some vector in . Then, you've just got to solve this pair of simultaneous equations for and :Does that make sense? (Let me know if anything sounds confusing -- happy to help.)

For part (b), suppose that instead of and you have and . This gives you the system:What sorts of conditions do you need here for that to have a solution?

Thanks phrontister, that work pretty well. Hard to believe it has been two whole years.

Anthony Lahmann wrote:

I will integrate 1/sqrt(1-x^2) by u-substitution. Here's how I did it:

After we did the u-substitution, we end up with the exact same integral, but with a negative in the front. What happened?

Your issue is here:

The correct implication is:

How many breakdowns occur in total?