How to prove that $(a^m)^n=a^{mn}$ where $a,m,n$ are real numbers and a>0? I know how to prove the equality when $m$ is a rational number and $n$ is an integer, but do not know how to go about proving this for real numbers. On a semi-related note, I was trying to prove this when both $m$ and $n$ are rational, and found out that I have to prove that $(\frac{1}{z})^{\frac{1}{y}}$=$\frac{1}{z^{\frac{1}{y}}}$. Does this need to be proven or can I accept it as a definition?

The very first thing you need to do is ask yourself what the definitions are. Without proper definitions, you'll never have a complete proof. So, if $a>0$ and $m\in \Bbb{R}$, how are you even supposed to define $a^m$? This is not at all a trivial task.
For example, here's one possible approach to things:

First, define $\exp: \Bbb{R} \to \Bbb{R}$ by $\exp(x):= \sum_{n=0}^{\infty}\frac{x^n}{n!}$. You of course have to check that this series converges for every $x\in \Bbb{R}$.
Check basic properties of $\exp$, such as $\exp(0) = 1$ and for all $x,y \in \Bbb{R},$ $\exp(x+y) = \exp(x)\cdot \exp(y)$. Also, verify that $\exp:\Bbb{R} \to (0,\infty)$ is an invertible function.
Since $\exp:\Bbb{R} \to (0,\infty)$ is invertible, we can consider its inverse function, which we denote as $\log:(0,\infty) \to \Bbb{R}$. Then, verify all the basic properties of $\log$, such as for all $a,b>0$, $\log(ab) = \log(a) + \log(b)$.
Finally, given $a>0$ and $m\in \Bbb{R}$, we define $a^m := \exp(m \log(a))$.

From this point, it is a simple matter to use the various properties of the exponential and logarithmic functions: for any $a>0$ and $m,n \in \Bbb{R}$
\begin{align}
a^{m+n} &:= \exp((m+n)\log (a)) \\
&= \exp[m\log(a) + n \log (a)]\\
&= \exp[m \log(a)] \cdot \exp[n\log(a)] \\
&:= a^m \cdot a^n \tag{$*$}
\end{align}
Similarly,
\begin{align}
(a^m)^n &:= \exp[n \log(a^m)] \\
&:= \exp[n \log(\exp(m \log(a)))] \\
&= \exp[nm \log(a)] \tag{since $\log \circ \exp = \text{id}_{\Bbb{R}}$} \\
&:= a^{nm} \\
&= a^{mn}
\end{align}
where in the last line, we make use of commutativity of multiplication of real numbers.

Note that steps 1,2,3 are not at all trivial, and indeed there are entire chapters of calculus/analysis textbooks devoted to proving these facts carefully. So, while I only listed out various statements, if you want the proofs for the statements I made, you should take a look at any analysis textbook, for example, Rudin's Principles of Mathematical Analysis, or Spivak's Calculus (I recall Spivak motivating these things pretty nicely).
As for your other question, yes it is something which needs to be proven. This result can be easily deduced from two other facts.

For any $x\in \Bbb{R}$, $1^x = 1$. (proof: $1^x := \exp[x \log(1)] = \exp[0] = 1$)
For any $a,b > 0$ and $x\in \Bbb{R}$, $(ab)^x = a^x b^x$. The proof is a few lines, once you use the properties of $\exp$ and $\log$.

Now, if $z>0$, then for any $x\in \Bbb{R}$,
\begin{align}
z^x \cdot \left(\frac{1}{z}\right)^x &= \left(z\cdot \frac{1}{z}\right)^x = 1^x = 1
\end{align}
Hence, $\left(\frac{1}{z}\right)^x = \frac{1}{z^x}$. In particular, you can take $x=1/y$ to prove what you wanted.

Edit: Motivating the definition $a^x := \exp(x\log(a))$, for $a>0, x \in \Bbb{R}$.
The long story short: this definition is unique in certain sense, and is almost forced upon us once we impose a few regularity conditions.
Now, let me once again stress that you should be careful to distinguish between definitions, theorems and motivation. Different authors have different starting points, so Author 1 may have one set definitions and motivations, and hence different theorems, while author 2 may have a completely different set of definitions, and hence have different theorems, and motivation.
So, let's start with some motivating remarks. Fix a number $a>0$. Then, we usually start start by defining $a^1 = a$. Next, given a positive integer $m\in \Bbb{N}$, we define $a^m = \underbrace{a\cdots a}_{\text{$m$ times}}$ (If you want to be super formal, then ok, this is actually a recursive definition: $a^1:= 1$, and then for any integer $m\geq 2$, we recursively define $a^{m}:= a\cdot a^{m-1}$).
Now, at this point what we observe from the definition is that for any positive integers $m,n\in \Bbb{N}$, we have $a^{m+n} = a^m \cdot a^n$. The proof of this fact follows very easily by induction.
Next, we typically define $a^0 = 1$. Why do we do this? One answer is that it is a definition, so we can do whatever we want. Another answer, is that we are almost forced to do so. Why? notice that for any $m\in \Bbb{N}$, we have $a^m = a^{m+0}$, so if we want this to be equal to $a^m \cdot a^0$, then we had better define $a^0 = 1$.
Next, if $m>0$ is an integer, then we usually define $a^{-m} := \dfrac{1}{a^{m}}$. Once again, this is just a definition, so we can do whatever we want. The motivation for making this definition is that we have $1 =: a^0 = a^{-m+m}$ for any positive integer $m$. So, if we want the RHS to equal $a^{-m}\cdot a^m$, then we had better define $a^{-m}:= \frac{1}{a^m}$.
Similarly, if $m>0$, then we define $a^{1/m} = \sqrt[m]{a}$ (assuming you've somehow proven existence of $m^{th}$ roots of positive real numbers). Again, this is just a definition. But why do we do this? Because we have $a =: a^1 = a^{\frac{1}{m} + \dots +\frac{1}{m}}$, so if we want the RHS to equal $(a^{\frac{1}{m}})^m$, then of course, we had better define $a^{1/m}:= \sqrt[m]{a}$.
Finally, we define $a^{\frac{m}{n}}$, for $m,n \in \Bbb{Z}$ and $n >0$ as $a^{m/n} = (a^{1/n})^m$. Once again, this is just a definition, so we can do whatever we want, but the reason we do this is to ensure the equality $a^{m/n} = a^{1/n + \dots + 1/n} = (a^{1/n})^m$ is true.
Now, let's think slightly for what we have done. We started with a number $a>0$, and we defined $a^1 := a$, and we managed to define $a^x$ for every rational number $x$, simply by the requirement that the equation $a^{x+y} = a^x a^y$ hold true for all rational $x,y$. So, if you actually read through everything once again, what we have actually done is shown the following theorem:

Given $a>0$, there exists a unique function $F_a:\Bbb{Q} \to \Bbb{R}$ such that $F_a(1) = a$, and such that for all $x,y\in \Bbb{Q}$, $F_a(x+y) = F_a(x)\cdot F_a(y)$.
(Note that rather than writing $a^x$, I'm just writing $F_a(x)$, just to mimic the function notation more)

Our motivation has actually been to preserve the functional equation $F_a(x+y) = F_a(x)\cdot F_a(y)$ as much as possible. Now, we can ask whether we can extend the domain from $\Bbb{Q}$ to $\Bbb{R}$, while preserving the functional equation, and if such an extension is unique. If the answer is yes, then we just define $a^x := F_a(x)$ for all real numbers $x$, and then we are happy.  It turns out that if we impose a continuity requirement, then the answer is yes; i.e the following theorem is true:

Given $a>0$, there exists a unique continuous function $F_a:\Bbb{R} \to \Bbb{R}$ such that $F_a(1) = a$, and such that for all $x,y\in \Bbb{R}$, $F_a(x+y) = F_a(x)\cdot F_a(y)$.

Uniqueness is pretty easy (because $\Bbb{Q}$ is dense in $\Bbb{R}$ and $F_a$ is continuous). The tough part is showing the existence of such an extension.
Of course, if you already know about the $\exp$ function and its basic properties like 1,2,3, then you'll see that the function $F_a:\Bbb{R} \to \Bbb{R}$ defined by $F_a(x):= \exp(x \ln(a))$ has all the nice properties (i.e is continuous, it satisfies that functional equation, and $F_a(1) = a$). Because of this existence and uniqueness result, this is the only reasonable way to define $a^x \equiv F_a(x) := \exp(x \log(a))$; anything other than this would be pretty absurd.
The purpose of the rest of my answer is to try to motivate how anyone could even come up with the function $F_a(x) = \exp(x\ln(a))$; sure the existence and uniqueness result is very nice and powerful, but how could you try to come up with it by yourself? This certainly doesn't come from thin air (though at some points we have to take certain leaps of faith, and then check that everything works out nicely).

To do this, let's start with a slightly more restrictive requirement. Let's try to find a function $f:\Bbb{R} \to \Bbb{R}$ with the following properties:

for all $x,y\in\Bbb{R}$, $f(x+y) = f(x)\cdot f(y)$
$f$ is non-zero; i.e there exists $x_0\in \Bbb{R}$ such that $f(x_0) \neq 0$.
$f$ is differentiable at $0$.

The first two conditions seem reasonable, but the third one may seem a little strange, but let's just impose it for now (it's mainly there to try to motivate things and hopefully simlify the argument and to convince you that $x\mapsto \exp(x\ln(a))$ didn't come from thin air).
First, we shall deduce some elementary consequences of properties 1,2,3:

In (2), we assumed $f$ is non-zero at a single point. We'll now show that $f$ is no-where vanishing, and that $f(0)=1$. Proof: we have for any $x\in\Bbb{R}$, $f(x) \cdot f(x_0-x) = f(x_0) \neq 0$. Hence, $f(x) \neq 0$. In particular, $f(0) = f(0+0) = f(0)^2$. Since $f(0)\neq 0$, we can divide it on both sides to deduce $f(0) = 1$.

We also have for every $x \in \Bbb{R}$, $f(x)>0$. Proof:  We have
\begin{align}
f(x) = f(x/2 + x/2) = f(x/2)\cdot f(x/2) = f(x/2)^2 > 0,
\end{align}
where the last step is because $f(x/2) \neq 0$ (this is why in real analysis, we always impose the condition $a = f(1) > 0$).

$f$ is actually differentiable on $\Bbb{R}$ (not just at the origin). This is because for $t\neq 0$, we have
\begin{align}
\dfrac{f(x+t) - f(x)}{t} &= \dfrac{f(x)\cdot f(t) - f(x) \cdot f(0)}{t} = f(x) \cdot \dfrac{f(0+t) - f(0)}{t}
\end{align}
now, the limit as $t\to 0$ exists by hypothesis since $f'(0)$ exists. This shows that $f'(x)$ exists and $f'(x) = f'(0) \cdot f(x)$. As a result of this, it immediately follows that $f$ is infinitely differentiable.

Now, we consider two cases. Case ($1$) is that $f'(0) = 0$. Then, we have $f'(x) = 0$ for all $x$, and hence $f$ is a constant function, $f(x) = f(0) = 1$ for all $x$. This is clearly not very interesting. We want a non-constant function with all these properties. So, let's assume in addition that $f'(0) \neq 0$. With this, we have that $f'(x) = f'(0)\cdot f(x)$; this is a product of a non-zero number and a strictly positive number. So, this means the derivative $f'$ always has the same sign. So, $f$ is either strictly increasing or strictly decreasing. Next, notice that $f''(x) = [f'(0)]^2 f(x)$, is always strictly positive; this coupled with $f(x+y) = f(x)f(y)$ implies that $f$ is injective and has image equal to $(0,\infty)$. i.e $f:\Bbb{R} \to (0,\infty)$ is bijective.

Theorem 1.

Let $f:\Bbb{R} \to \Bbb{R}$ be a function such that:

for all $x,y\in \Bbb{R}$, $f(x+y) = f(x)f(y)$
$f$ is non-zero
$f$ is differentiable at the origin, with $f'(0) \neq 0$

Suppose $g:\Bbb{R} \to \Bbb{R}$ is a function which also satisfies all these properties. Then, there exists a number $c\in \Bbb{R}$ such that for all $x\in \Bbb{R}$, $g(x) = f(cx)$. In other words, such functions are uniquely determined by a constant $c$.
Conversely, for any non-zero $c\in \Bbb{R}$, the function $x\mapsto f(cx)$ satisfies the three properties above.

Proof
To prove this, we use a standard trick: notice that
\begin{align}
\dfrac{d}{dx}\dfrac{g(x)}{f(cx)} &= \dfrac{f(cx) g'(x) - g(x) cf'(cx)}{[f(cx)]^2} \\
&= \dfrac{f(cx) g'(0) g(x) - g(x) c f'(0) f(cx)}{[f(cx)]^2} \\
&= \dfrac{g'(0) - c f'(0)}{f(cx)} \cdot g(x)
\end{align}
Therefore, if we choose $c = \dfrac{f'(0)}{g'(0)}$, then the derivative of the function on the LHS is always zero. Therefore, it must be a constant. To evaluate the constant, plug in $x=0$, and you'll see the constant is $1$. Thus, $g(x) = f(cx)$, where $c= \frac{g'(0)}{f'(0)}$. This completes the proof of the forward direction. The converse is almost obvious
Remark
Notice also that from $g(x) = f(cx)$, by plugging in $x=1$, we get $g(1) = f(c)$, and hence $c = (f^{-1} \circ g)(1) = \frac{g'(0)}{f'(0)}$ (recall that we already stated that such functions are invertible from $\Bbb{R} \to (0,\infty)$). It is this relation $c = (f^{-1} \circ g)(1)$, which is the key to understanding where $x\mapsto \exp(x\ln(a))$ comes from. We're almost there.

Now, once again, just recall that we have been assuming the existence of a function $f$ with all these properties. We haven't proven the existence yet. Now, how do we go about trying to find such a function $f$? Well, recall that we have the fundamental differential equation $f'(x) = f'(0) f(x)$. From this, it follows that for every positive integer $n$, $f^{(n)}(0) = [f'(0)]^n$. We may WLOG suppose that $f'(0) = 1$ (other wise consider the function $x\mapsto f\left(\frac{x}{f'(0)}\right)$), then we get $f^{(n)}(0) = 1$. Finally, if we make the leap of faith that our function $f$ (which is initially assumed is only differentiable at $0$ with $f'(0) = 1$, and then proved it is $C^{\infty}$ on $\Bbb{R}$) is actually analytic on $\Bbb{R}$, then we know that the function $f$ must equal its Taylor series:
\begin{align}
f(x) &= \sum_{n=0}^{\infty} \dfrac{f^{(n)}(0)}{n!} x^n = \sum_{n=0}^{\infty}\dfrac{x^n}{n!}
\end{align}
This is one of the many ways of how one might guess the form of the exponential function, $\exp$. So, we now take this as a definition: $\exp(x):= \sum_{n=0}^{\infty}\frac{x^n}{n!}$. Of course using basic power series techniques, we can show that $\exp$ is differentiable everywhere, and satisfies that functional equation with $\exp(0)=\exp'(0) = 1$.
So, now, back to our original problem. Given any $a>0$, we initially wanted to find a function $F_a:\Bbb{R} \to \Bbb{R}$ such that $F_a$ satisfies the functional equation, and $F_a(1) = a$, and such that $F_a$ is differentiable at $0$ with $F_a'(0) \neq 0$. Well, in this case, both $F_a$ and $\exp$ satisfy the hypothesis of theorem 1. Thus, there exists a constant $c \in \Bbb{R}$ such that for all $x\in \Bbb{R}$, $F_a(x) = \exp(cx)$. To evaluate the constant $c$, we just plug in $x=1$, to get $c = (\exp^{-1}\circ F_a)(1) := \log(a)$. Therefore we get $F_a(x) = \exp(x \log(a))$. This is why we come up with the definition $a^x := \exp(x\log(a))$.

The rule is merely a specific case of the addition law of indices/powers. Consider:
$$(a^m)^n=a^m\times a^m \times a^m \times a^m \times a^m...\times a^m$$
where there are $n$ lots of $a^m$ multiplied together. We have the law that $$a^p \times a^q=a^{p+q}$$
Applying this to your question we have:
$$(a^m)^n=a^{m+m+m+m+m...+m}$$
where we have  $n$ lots of $m$. Now, $n$ lots of $m$ is equal to $n\times m$ which is equal to $mn$.
So, finally, we have that
$$(a^m)^n=a^{mn}$$
as required.
Edit: Since writing the above proof for positive integer exponents I have been shown a proof that $\ln {a^x}=x\ln a $ without using what we are trying to prove in the question itself, which allows us to use the laws of logs etc to answer the question, as previous answers have done:
Consider $$f(x)=\ln {x^n} -n\ln x\implies f'(x)=\frac{nx^{n-1}}{x^n}-\frac{n}{x}=\frac{n}{x}-\frac{n}{x}=0$$
This means that $f(x)$ must be equal to some constant, $c$, as only constants diiferentiate to $0$. Let's try to find $c$: We have
$$\ln {x^n} -n\ln x=c$$
Let $x=1$:
$$\ln1-n\ln1=c=0-0=0$$
So we have $c=0$, leaving us with $$\ln {x^n} =n\ln x$$
For the sake of completeness, I'll now go on to answer the question for all real exponents. Note that $\ln {x^n} =n\ln x$ is true for all $n\in\mathbb R$.
Let $x=a^m$:
$$e^{\ln {x^n}}=e^{\ln {(a^m)^n}}=(a^m)^n$$
But we also have
$$e^{\ln {x^n}}=e^{n\ln {x}}=e^{n\ln {a^m}}=e^{mn\ln {a}}=e^{\ln {a^{mn}}}=a^{mn}$$
So, at last, we have:
$$(a^m)^n=a^{mn}$$.
Thanks to peek-a-boo, among others, who attempted to make me understand this in more detail.


Questions regarding The Fundamental Theorem of Calculus FTC1 suggested that
$$
F'(x) = \frac{d}{dx}\int_a^xf(t)dt = f(x)
$$
and the Chain Rule says that
$$
\frac{\mathrm{d}}{\mathrm{d}x}F\big(g(x)\big)=F'\big(g(x)\big)·g'(x).
$$
That leads to
$$
\frac{\mathrm{d}}{\mathrm{d}x}F\big(g(x)\big) = g'(x)·\frac{\mathrm{d}}{\mathrm{d}g(x)} \int\limits_a^{g(x)}\!\!f(t)\,\mathrm{d}t = f\big(g(x)\big)·g'(x)
$$
Is it right to combine these two?

If it is, what should I do if I want to find $F'(x)$ instead of $\frac{\mathrm{d}}{\mathrm{d}x}F\big(g(x)\big)$ in the combined equation?

I'm actually confused in the following question:

f(t) = $ \int\limits_2^t( {\sqrt {\frac{7}{4}+u^3}}) du  $

F(x) = $ \int\limits_1^{\sin x}f(t)dt $

Find: $F''(\pi)$

Leibniz integral rule is what you need

$$\frac d{dx}F(x)=\frac d{dx}\int_{g(x)}^{h(x)}f(t)dt =f(h(x))h'(x) - f(g(x))g'(x)$$
$$F'(x)=f(\sin x)\cdot\cos(x)=\cos(x)\int_2^{\sin(x)}\sqrt{u^3+7/4}\ \ du$$
Can you take it from here.

PS: I see you are trying to derive this formula. Look at the wikipedia page for the proofs.

Hint: You know how to differentiate $F(g(x)).$ You want to differentiate $F(x).$ That's just the first one when $g(x)=x.$

I see you've seen that without sufficient context, you may be unclear about what you want, therefore misleading potential answerers about your real problem. Here, instead of being all vague about arbitrary functions, you should have noted right from the get go what it was that was bothering you about a specific question. You only added this later after probably realising that no one seems to understand what you mean. I hope this is a lesson for next time.

In any case, to explain your difficulty, you want to find $F''(π),$ where $$F(x)=\int_1^{\sin x}{f(t)\mathrm d t},$$ and $$f(t)=\int_2^t{\sqrt{\frac74 + u^3}\mathrm d u}.$$

Now, $$F'(x)=(\sin x)'f(\sin x)=\cos x\int_2^{\sin x}{\sqrt{\frac74 + u^3}\mathrm d u}.$$ Thus, we have that $$F''(x)=(\cos x)'\int_2^{\sin x}{\sqrt{\frac74 + u^3}\mathrm d u}+(\cos x)(\sin x)'\sqrt{\frac74 + \sin^3x}=-\sin x\int_2^{\sin x}{\sqrt{\frac74 + u^3}\mathrm d u}+(\cos x)^2\sqrt{\frac74 + \sin^3x}.$$ Now set $x=π.$


Probability of Running Around Laps The question is as follows:

  Jack is supposed to run laps around the outdoor track. At the start of each lap, including the first, there is always an $8$ percent chance that Jack will not run for the day. What is the probability that Jack will run (a) no laps? (b) at least four laps? (c) exactly four laps?

(a)
 The chances of running no laps is $8$%, or $\frac{2}{25}$.

(b) The chances of running at least four laps is $(\frac{23}{25})^4$.

I am not completely sure about (c), especially in making sure whether he will run the exact four laps.

If there are any errors, please let me know. Any help will be greatly appreciated.

You are correct so far.  For c, he needs to run at least four laps, which is your answer to b, and then run no more laps, which is your answer to a.

In order to run exactly 4 laps he must run

a) the first lap which has probability, as you say, 2/25

b) the second lap which also has probability 2/25

c) the third lap which also has probability 2/25

d) the fourth lap which also has probability 2/25

e) NOT run the fifth lap which has probability 23/25.

The probability of running exactly 4 laps is $\left(\frac{2}{25}\right)^4\left(\frac{23}{25}\right)$


Limit of integration can't be the same as variable of integration? I am told that an expression like
$$
\int_a^x f(x)dx
$$
is not well formed, i.e. it should be
$$
\int_a^xf(t)dt
$$
or similar.

Why is it that the limits of integration can't depend on the variable of integration?

In mathematics, it's generally regarded as a bad idea for the same symbol to have two different meanings in the same expression. In this case, the variable being integrated with respect to effectively disappears, and a new variable (really two new variables, the bounds of integration) takes over. To call them the same thing can make things confusing sometimes (although not always). This is more of a stylistic than a strictly logical concern, at least in one variable.

your question is very valid and I see all the answers posted say a very comfortable no. I can give a physical situation where the limit is also a function of the variable in the integral (one which I am trying to solve myself). It goes like this:

The solar radiation intensity ($W/m^2$) at a point on the surface of the earth depends on the angle (theta) between the incident ray and the normal (radius vector) at the point. The exercise is to find the total radiant energy ($J/m^2$) over the period of a year. Now, theta as a function of time varies continuously and the limit of integration would cover the longitudinal sweep from sunrise to sunset. Bear in mind that this longitudinal sweep of daylight (from sunrise to sunset) varies with latitude of the point and also the time of year; thus the daily limit of integration is also a function of time (and this has to be summed up over a year).

I hope the problem statement is clear and I have worked out the angle relationships between theta, latitude, tilt of earth axis and angle swept by the earth in its orbit around the sun (taking the northern hemisphere summer solstice as the starting point).

It is a very practical problem and I'm sure that a solution to this exists that may have been discovered or is yet to be discovered.


Application of Jensen's Inequality-Positive Definite Matrices-Probability I'm given the question for $A$ positive definite matrix in $\mathbb{R}^n$, use Jensen's inequality to prove $(x^TAx)(x^TA^{-1}x)\geq1$ for unit vectors $x\in\mathbb{R}^n$. The hint is to think about $A$ diagonal and then diagonalize $A$ in the general case. (If $A$ is diagonal, $(x^TAx)(x^TA^{-1}x)=1$, but positive definite does not imply diagonalizable).

I know $x^TAx>0$ for all $x\in\mathbb{R}^n$. But in order to use Jensen's inequality, I need to have a convex function. Here is what I have thought about:

Let $g(y)=1/y$ is a convex function, then $g(\mathbb{E}y)\leq\mathbb{E}(g(y))$. I want then to get $y=(x^TAx)(x^TA^{-1}x)$ to see $\frac{1}{(x^TAx)(x^TA^{-1}x)}\leq 1$. I know this does not work, but it is how I am trying to use Jensen.

Hint: If $D$ is diagonal with elements $d_1,d_2\ldots, d_n$, then $D^{-1}$ is diagonal with elements $\frac1{d_1},\ldots,\frac1{d_n}$. Then
$$x^TDx = \sum_i d_ix_i^2 = E(Y)$$
where $Y$ takes value $d_i$ with probability $x_i^2$. (Note that $x$ has norm 1, so the probabilities add up to 1.) What is the corresponding statement for $x^TD^{-1}x$? Now apply Jensen's inequality $Ef(Y)\ge f(E(Y))$ with a suitable choice of $f$. Spoiler below:

   $f(y)=\frac1y$.

Fix $x$, the map $B\xrightarrow{\quad\pi\quad}\left\langle x,Bx\right\rangle $
is positive definite on $M\left(n,\mathbb{C}\right)$. So by Cauchy-Schwarz inequality,

\begin{equation}
\left|\pi\left(A^{1/2}A^{-1/2}\right)\right|^{2}\leq\pi\left(A^{1/2}A^{1/2}\right)\pi\left(A^{-1/2}A^{-1/2}\right),
\end{equation} i.e.,  $\left\langle x,Ax\right\rangle \left\langle x,A^{-1}x\right\rangle \geq1$, with $\left\Vert x\right\Vert =1$.


projective geometry and projective space Let $V$ is a vectorspace  over field $F_q$, we denote the set of all subspaces of $V$ by $\mathcal{P}(V)$. I saw some referencess they mentioned $\mathcal{P}(V)$ as a projective space and some referencess they mentioned as projective geometry. what is the difference between a projective geometry and a projective space?

Geometry is defined as structure $G=(\Omega,I)$ consisting of set $\Omega$ and a relation I. That definition is pretty abstract and can be used on whole variety of objects and relations.
Once you get points and lines as elements of $\Omega$ and incidence relation for I, you get projective geometry.

Now we introduce axioms of projective geometry (which I am not going to write here) and we can define projective space as geometry $G=(\Omega,I)$ which satisfies given axioms.

Beutelspacher, Albrecht; Rosenbaum, Ute, Projective geometry. From foundations to applications, Vieweg Studium 41. Aufbaukurs Mathematik. Braunschweig: Vieweg (ISBN 3-528-17241-X/pbk). x, 265 p. (2004). ZBL1050.51001.

Projective geometry is the study of projective spaces.


Find $\lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$ without L'Hopital or Taylor series. $$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$$

My try is as follows:

$$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=$$$$ \lim_{x\to
∞}x\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$$$=\lim_{x\to ∞}x\lim_{x\to
∞}\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$
which is $∞×0$ , but clearly this zero is not exactly zero. I was thinking about generalized binomial
theorem, but seems it will make the limit difficult, so how this kind of limits can be solved without using Taylor series or L'Hopital's rule?

We first note that for any positive integer $n$ and any real $a$,
$$\lim_{x\to \infty}x\left(\sqrt[n]{1+\frac{a}{x}}-1\right)=
\lim_{s\to 1}a\frac{s-1}{s^n-1}=\lim_{s\to 1}\frac{a}{s^{n-1}+s^{n-2}+\dots +s +1}=\frac{a}{n}$$
where $s=\sqrt[n]{1+a/x}$ and therefore $a/x=s^n-1$, and $x=a/(s^n-1)$.

Hence, from your work, we split the limit in two:
$$\begin{align}\lim_{x\to +\infty} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})
&=\lim_{x\to +\infty}x\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)
\\&=\lim_{x\to +\infty}x\left(\sqrt[3]{1+\frac{3}{x}}-1\right)-\lim_{x\to \infty}x\left(\sqrt[2]{1 +\frac{-2}{x}}-1\right)\\&=\frac{3}{3}-\frac{-2}{2}=1+1=2.
\end{align}$$

P.S. Note that on the other side,
$$\lim_{x\to -\infty} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=-\infty$$

From OP's work
$$L=\lim_{x \rightarrow \infty} x \left[ \left( 1+\frac{3}{x} \right)^{1/3}-\left( 1-\frac{2}{x} \right)^{1/2}\right]= \lim_{x \rightarrow \infty} x \left[ \left( 1+\frac{3}{3x} \right)-\left( 1-\frac{2}{2x} \right)\right]= \lim_{x \rightarrow \infty}x \frac{2}{x}=2. $$


Is there any sequence of functions in $C_c (\mathbb R)$ which converges pointwise only to $e^{-x^2}$ This question has been posted earlier.  I could not understand the solution. Can anyone please help me to understand the solution?

let $C_c(\mathbb{R})$ = { f : $\mathbb{R} \rightarrow \mathbb{R}$ | f is continuous  and there  exist  a  compact  set $K$ such that  $f(x) = 0$ for all $x \in K ^c$} . let $g(x) = e^{-x^2}$  for all $x \in$ $\mathbb{R}.$

which of the  following satemnet is true?

1.There  exist a sequence {$f_n$} in $C_c(\mathbb{R})$ such that $f_n \rightarrow g$ uniformly

2.There exist a sequemce  {$f_n$} in $C_c (\mathbb{R})$ such that $f_n \rightarrow g$ pointwise

If  a sequence in $C_c (\mathbb{R})$ converge pointwise to g then it must converge uniformly to $g$.

4.There doesnot  exists any sequence $C_c (\mathbb{R})$ converging pointwise to $g$

I can not understand the solution of 3. Can anyone please make me understand?

False. Start from $f_n$ but add a function $\psi_n\in C_c^{\infty}(\mathbb{R})$ with $\operatorname{supp}\psi_n\subset [n+1,n+2]$, and such that $\psi_n(\xi)=-1$ for some $\xi\in (n+1,n+2)$. Then $g_n:=f_n+\psi_n\in C_c^{\infty}(\mathbb{R})$ and $g_n(x)\to e^{-x^2}$ for all $x\in \mathbb{R}$, but $g_n$ does not converge uniformly to $e^{-x^2}$ because
$$\sup_{x\in \mathbb{R}}|g_n(x)-e^{-x^2}|\geq |g_n(\xi)-e^{-\xi^2}|\geq 1  $$

The point is that you can take a sequence, even one that converges uniformly to $g$, and tweak it in such a way that it still converges pointwise, but not uniformly.
Consider first $g=0$. And take $g_n$ to be a "traveling bump". For instance,
$$
g_n(t)=\begin{cases}
0,&\ t\not\in[n,n+1]\\ \ \\
2t-2n,&\ t\in [n,n+1/2]\\ \ \\
-2t+2n+2,&\ t\in [n+1/2,n+1]
\end{cases}
$$
For any fix $t$, if $n>t$ then $g_n(t)=0$. So $g_n\to0$ pointwise. But not uniformly, because $$g_n(n+1/2)-g(n+1/2)=1$$ for all $n$.
The solution you were given is basically to add the above example to a sequence $f_n$ converging to $e^{-x^2}$.

Quick answer: For Q.1 and Q.2, Yes. The following is a well known fact in general topology and functional analysis.

For any locally compact Hausdorff space $\Omega$, the set $C_c(\Omega)$ is a linear subspace of $C_0(\Omega)$. Moreover, $C_c(\Omega)$ is norm-dense in $C_0(\Omega)$ with respect to the supremum norm.

For your particular case, $\mathbb{R}$ is a locally compact Hausdorff topological space with respect to the usual topology and $x\mapsto \exp(-x^2)$ is an element in $C_0(\mathbb{R})$. The results follows immediately.


Sum of coefficients of $x^i$ (Multinomial theorem application) A polynomial in $x$ is defined by
  $$a_0+a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}=(x+2x^2+ \cdots +nx^n)^2.$$
  Show that the sum of all $a_i$, for $i\in\{n+1,n+2, \ldots , 2n\}$, is
  $$ \frac {n(n+1)(5n^2+5n+2)} {24}.$$

I don't know how to proceed. I know the Multinomial theorem, however, I have problems in applying it. Any help will be appreciated as it will help me understand the theorem well.

Thanks!

Here is an easy method using multinomial coefficients. Put $x=1$ to get the sum of all the coefficients. Now, we want to evaluate $\sum_{i=0}^na_i$, then we will subtract that from sum of all coefficients. Observe that these coefficients will remain unaltered even in the following expansion (because the additional terms do not contribute to powers less than $x^{n+1}$):
$$(x+2x^2+3x^3+...)^2 = x^2(1+2x+3x^2+...)^2$$
$$ = x^2\Bigg(\frac{1}{(1-x)^2}\Bigg)^2$$
$$ = \frac{x^2}{(1-x)^4}$$
$$ = x^2\sum_{m=0}^\infty\binom{m+4-1}{4-1}x^m$$
Now, apply the identity that $$\sum_{i=k}^n\binom{i}{k} = \binom{n+1}{k+1}$$
and you are done.

Hope it helps:)

You can check in the link it's similar to what your question is.
https://www.mathsdiscussion.com/forum/topic/sum-of-coefficients/?part=1#postid-55


You have the right answer; you just have a different constant. Using $C_1$ instead of $C$, set your answer
$$\frac{x^2-3}{2}+x+\ln|x-1|+C_1$$
equal to option (C) and simplify. You'll get $C-C_1=-\frac32$, which is fine since the difference is constant.

Now, as an alternative approach to the problem, consider that you could find the derivative of each option and see which one reduces to $\frac{x^2}{x-1}$. Differentiation is typically easier than integration, and with multiple-choice questions it can be helpful to work backwards.

As a side note, it isn't really correct to say that $\int\frac1x\ dx=\ln|x|+C$, because of the discontinuity of $\frac1x$ at $0$. It's a bit more nuanced than that:
$$
\int\frac1x\ dx =
\begin{cases}
\ln|x| + C_1, & \text{if $x > 0$} \\
\ln|x| + C_2, & \text{if $x < 0$}
\end{cases}$$

So in that sense, even the given answers are not entirely right.

After some rather basic algebraic manipulations, all you're left with is a pretty simple u-substantiation problem:

$$
\begin{align}
\int \frac{x^2}{x-1} \,dx
&=\int \frac{x^2-1+1}{x-1} \,dx\\
&=\int \left(\frac{x^2-1}{x-1}+\frac{1}{x-1}\right) \,dx\\
&=\int \frac{(x-1)(x+1)}{x-1}\,dx+\int\frac{1}{x-1} \,dx\\
&=\int \left(x+1\right)\,dx+\int\frac{1}{x-1}\frac{d}{dx}\left(x-1\right) \,dx\\
&=\frac{x^2}{2}+x+\int\frac{1}{x-1}\,d\left(x-1\right)\\
&=\frac{x^2}{2}+x+\ln{\left|x-1\right|}+C.\\
\end{align}
$$


Find the point, if any, the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line Section 2.5 #14

Find the point, if any, the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line.

Okay, so having a horizontal tangent line at a point on the graph means that the slope of that tangent line is zero. The derivative of a function is another function that tells us the slope of the tangent line at any given point on the graph of the original function.

Thus, to find where the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line, we need to take the derivative, set it equal to zero, and solve for $x$. This will give us the $x$-coordinate of where the graph of $f(x)$ has a horizontal tangent line. To find the corresponding $y$ value, we plug the $x$ value that we found into the original equation.

In this problem, when we plug the $x$ value we find into the original equation, we get an imaginary number, which means that no point on the graph of $f(x)$ has a horizontal tangent line, and thus our answer is DNE, does not exist. Let's go through the motions!!!

$f(x) = \sqrt{8x^2+x-3}=(8x^2+x-3)^{1/2}$

$f'(x) = \frac{d}{dx}(8x^2+x-3)^{1/2}$

Time do the chain rule!!!

$$\begin{align}
f'(x) &= \frac{(8x^2+x-3)^{-1/2}}{2}\frac{d}{dx}(8x^2+x-3)\\
&= \frac{(8x^2+x-3)^{-1/2}}{2}(16x+1)\\
&= \frac{(16x+1)}{2(8x^2+x-3)^{1/2}}
\end{align}$$

Alright, we have our derivative. We want to find horizontal tangent lines, so we set this equal to zero and solve for $x$

$$0 = \frac{(16x+1)}{2(8x^2+x-3)^{1/2}}$$

multiplying both sides of the equation by $2(8x^2+x-3)^{1/2}$ we get

$0 = (16x+1)$

And thus $x = \frac{-1}{16}$

Now, we plug this value into the original equation to get the corresponding $y$ value, because remember, we are looking for a point on the graph where the horizontal line is tangent, so our answer will be in $(x,y)$ format, is it exists, (which in this case, it won't)..

$f(\frac{-1}{16}) = \sqrt{8(\frac{-1}{16})^2+\frac{-1}{16}-3}$

But $8(\frac{-1}{16})^2+\frac{-1}{16}-3<0$, so taking its square root will give us an imaginary number. Thus the answer is DNE

Given

$$f(x) = \sqrt{8x^2+x-3}$$

the horizontal tangent lines of $f(x)$ occur all at points in the domain of $f(x)$ where $f'(x)=0$. You have correctly found the derivative

$$f'(x)=\frac{16x+1}{2\sqrt{8x^2+x-3}}$$

which is a rational function. The zeroes of a rational function are where the numerator is zero. Solving $16x+1=0$, we find that $x=-1/16.$

In order to have a horizontal tangent line at $x=-1/16$, $f(x)$ must be defined at $x=-1/16$.
The domain of $f(x)=\sqrt{8x^2+x-3}$ is where $8x^2+x-3\ge0$. Solving this inequality, we find the domain of $f(x)$ as

$$x\ge \frac{1}{16}\left(\sqrt{97}-1\right)$$
$$x\le \frac{1}{16}\left(-1-\sqrt{97}\right)$$

therefore since $$\frac{-1-\sqrt{97}}{16} < \frac{-1}{16}$$

we see that $x=-1/16$ is not in the domain of $f(x)$. So, $f(x)$ doesn't have any horizontal tangent lines.

First off, I assume you are working in $\mathbb{R}$, yes?  What is the domain of your original function?  Since $$f\left(x\right)=\sqrt{8x^{2}+x-3}\text{, the domain is implied: }8x^{2}+x-3\ge 0.$$Solve this to find the domain.  Then, when you find the $x$-value of the point where the tangent is $0$, see if that point is in the domain and if it is, you should end up with a real $y$-value, i.e., $y\in\mathbb{R}$.


For a given non-constant polynomial $f(x)$ with integer coefficients, how many solutions are there to $f(x)\equiv 0 \mod(n)$ where $n$ is composite? For a given non-constant polynomial $f(x)$ with integer coefficients, how many solutions are there to $f(x)\equiv 0 \mod(n)$ where $n$ is composite?

Is there a general way to determine the number of incongruent solutions modulo $n$?

My first idea is that we can of course break $n$ into its prime power factorization and look at $f(x)\equiv 0 \mod(p_{i}^{e_{i}})$ where $(p_{i}^{e_{i}})$ appears as a prime power factor in $n$.

Here's where I start to become confused, if $f(x)=x$ then the Chinese remainder theorem tells us that the solution is unique modulo $n$, but if $f(x)$ is non-constant and non-linear then we need to use the lifting method to solve $f(x)\equiv 0$ for each $\mod(p_{i}^{e_{i}})$ - but so far the method tells us nothing about the number of solutions.

I presume I am not incorrect in saying that the number of incongruent solutions to $f(x)\equiv \mod(p_{i}^{e_{i}})$ is at most $min(deg(f), p_{i}^{e_{i}})$, but is there a general way to determine precisely how many solutions are there?

First factor $m$ to powers of primes. $$N(m) = PI\ N(p^a)$$ $N$ is number of roots. This is by CRT.

Then for each prime factor p of m, a polynomial $P(x)$ can be reduced to $P'(x)$ with degree $<= p$. If $P'(x)$ is a factor of $(x^p - x) \mod p$ (after dividing, the remainder is multiple of $p$), then it has exactly $n$ roots where $n$ is degree of $P'(x)$.

From $p$ to $p^a$, it's Hensel's Lemma.


What is the double sum of: $\sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty \frac{ \sin[ka(m-n)]}{(m-n)} , m \neq n $
where $k$ and $a$ are constants.
How to treat this double sum?

Hint: Use the fact that $$\frac{\sin[(m-n)ka]}{m-n} = \frac 1 2\int_{-ka}^{ka}e^{i(m-n)t}dt$$

This is what i got.
If there is any error, let me know, please.


Can we calculate $2^k$ using this easy Taylor series? Trying to calculate $2^k$ by hand for $k\in[0,1]$, it's tempting to use the Taylor expansion of $x^k$ around $x=1$, to get:
$$2^k = 1^k + \frac{k (1)^{k-1}}{1!} + \frac{k(k-1) (1)^{k-2}}{2!} + \ldots =1+k +\frac{k(k-1)}{2!}+\ldots =\sum_{n=0}^\infty\binom{k}{n}$$
Unfortunately, $2$ lies exactly on the the radius of convergence $r = 1$, so in theory this may not converge.

Can we prove this converges to the correct value for all $k$? Numerically it does seem to.
What can be said about the rate of convergence? It seems quite slow. Can we bound the convergence?

Regarding convergence of the series, the $n$th term is

$$a_n = \frac{k(k-1) \ldots (k - n +1)}{n!}.$$

We have

$$\frac{a_{n+1}}{a_n} = \frac{k-n}{n+1} = - \frac{n-k}{n+1} < 0,$$

and the series is alternating for $n > k.$

Note that

$$\frac{|a_{n}|}{|a_{n+1}|} = \frac{n+1}{n-k} = \frac{1+1/n}{1-k/n} = 1 + \frac{1+k}{n} +O\left(\frac1{n^2}\right),$$

and

$$\lim_{n \to \infty} \left(n \frac{|a_n|}{|a_{n+1}|}- (n+1)\right) = k > 0.$$

There exists $N \in \mathbb{N}$ such that for $n > N$

$$n \frac{|a_n|}{|a_{n+1}|}- (n+1) > \frac{k}{2} \\ \implies |a_{n+1}| < \frac{2}{k}\left(n|a_n| - (n+1)|a_{n+1}|\right).$$

Thus for all $m > N$, the RHS forms a telescoping sum and

$$\sum_{n = N}^m |a_{n+1}| < \frac{2}{k}\left(N|a_N| - (m+1)|a_{m+1}|\right) < \frac{2}{k}N|a_N|.$$

The series $\sum|a_n|$ is positive and bounded, and, hence, convergent.

Therefore, the series $\sum_{n=0}^\infty\binom{k}{n}$ is absolutely convergent for $k > 0$.

As an alternating series an error bound is

$$\left|\sum_{n=m+1}^\infty\binom{k}{n}\right| \leqslant \left|\binom{k}{m+1}\right|.$$

Since $\left( \begin{array}{c} k \\ n \end{array}\right)=0$ for $n > k$ the series on the left side is always a finite sum. So to compute $2^k$ exact we need $k+1$ terms of this sum. So you can not speak of some sort of convergence.


Some Results in $\mathbb{Z} [\sqrt{10}]$ This is a question from an old Oxford undergrad paper on calculations in $\mathbb{Z} [\sqrt{10}]$. We equip this ring with the Eucliden function $d(a+b\sqrt{10})=|a^2-10b^2|$. I want to prove the following results:

If $d(x)=1$, then $\frac{1}{x} \in \mathbb{Z} [\sqrt{10}]$
Any non-zero element of $\mathbb{Z} [\sqrt{10}]$ which is not a unit can be expressed as a product of finitely many irreducibles in $\mathbb{Z} [\sqrt{10}]$
The ideal generated by $2$ and $\sqrt{10}$ is not principal in $\mathbb{Z} [\sqrt{10}]$

Thought so far

Suppose $x=a+b\sqrt{10}$. Clearly if $x$ is a unit then $d(x)=1$, though I'm not sure if this helps. Are we OK simply to note that $\frac{1}{x}=\frac{a-b\sqrt{10}}{a^2-10b^2}$ and since $d(x)=1$ then the deonminator is either $1$ or $-1$.
I know this is true in general in a principal ideal domain and every Euclidean ring is a principal ideal domain, but this proof is lengthy. Is there any calculation one can perform in $\mathbb{Z} [\sqrt{10}]$ to demonstrate this property more quickly.
Any help would be appreciated; I'm not actually too sure what this ideal looks set. Could someone put it in a set notation for me?

Many thanks.

It seems to me that (1) and (3) are dealt with adequately in the comments (but I would be happy to incorporate that here if you need). For (2), the key point is that the ring is Noetherian, which is effectively a consequence of the Hilbert basis theorem, implying that a polynomial ring $\mathbb{Z}[x]$ is Noetherian, and the fact that a quotient ring of a Noetherian ring is Noetherian. Given this, every element can be written as a product of irreducibles, since otherwise you obtain an infinite ascending chain of ideals.

Here are some more details for (3): First, observe that $d(xy)=d(x)d(y)$ for all $x,y$ and that $d(2)=4$ and $d(\sqrt{10})=10$. Thus any common divisor $x$ has $d(x)|2$. If $x$ is a non-unit this implies $d(x)=2$. Thus assuming a non-unit common divisor $x$ exists, there are integers $a,b$ with
$$\pm 2=a^2-10b^2.$$ Consider this equation modulo $10$. It implies that either $2$ or $8$ is a square mod $10$, but the squares modulo $10$ are just $0,1,4,9,6,5$. Hence there is no non-unit common divisor $x$. On the other hand, the ideal is proper since its elements are all of the form $a+b\sqrt{10}$ with $a$ even. Therefore it is not principal.

The set Z[10] makes use of half-primes, which must occur in even numbers.  This is pretty normal for many of these sorts of systems.  These half-primes live in some other part that have this same ratio, ie $a\sqrt{5}+b\sqrt{2}$.  In some systems, like z[3], there are even uits in there, eg $(\sqrt{6}+\sqrt{2})/2$.

In the case of Z[10] (and many other schemes), there are subspaces that hold primes, but there is no unit to move to that space.

An example is that of Z[210], which has three subspaces $a\sqrt{3}+b\sqrt{70}$ and $a\sqrt{2}+b\sqrt{105}$ and $a\sqrt{6}+b\sqrt{35}$, and a unit $\sqrt{15}+\sqrt{14}$, which transitions to another set of subspaces.  This group has no fewer than eight sub-spaces, linked in pairs, and primes can be individually found in any of the main group, or the three subgroups (3), (2), (6).  A number appears in the main group, if the product of the special primes of the subgroup, multiply up to a square.  For example the prime factors for 29 are $\sqrt{35}+\sqrt{6}$. and for 7, one can find it in $\sqrt{105}+7\sqrt{2}= \sqrt{7}*(\sqrt{15}+\sqrt{14})$, but since 6*2 gives 12, one needs a further '3' type prime to make it appear in Z[210].

That is pretty much the order of the day for composite numbers in this type.


Which is right way to calculate percentage? A student gets the following marks.

50   out of 100

120  out of 150

30   out of 50

In first method : I calculate the percentage as (sum of obtained marks) / (Total marks) * 100.
Hence [(50 + 120 + 30) / (100 + 150 + 50)] * 100
     = (200 / 300) * 100
     = 66.66%

In second method : I calculate individual percentages and divide by three as:

(50% + 80% + 60%) / 3 = (190 / 3) = 63.33%

Why are the two percentages different and which one is the correct percentage?

First notice that the two results are close together, although you can probably create some strange case in which they aren't close.
Second there is no "right" answer.  Whoever creates the scoring scheme can decide how to combine the marks.
Third the two answers are the result of two different weighting schemes.  There are infinitely many such schemes.

Your first method would be correct. The cause for the difference between the two methods is that the second one doesn't take into account the differing amounts of total marks between assignments. This leads to an unequal weighting between assignments that are worth more or less than others. For an extreme example, say a student go the following marks:

2 out of 2 (100%)
20 out of 100 (20%)

The first method would give us a correct answer of $\frac{2+20}{2+100} \approx 21.57\%$.

The second method gives us an incorrect answer of $\frac{100+20}{2} = 60\%$


Do most mathematicians know most topics in mathematics? How many topics outside of his or her specialization is an average mathematician familiar with?

For example, does an average group theorist know enough of partial differential equations to pass a test in a graduate-level PDE course?

Also, what are the "must-know" topics for any aspiring mathematician? Why?

As a graduate student, should I focus more on breadth (choosing a wide range of classes that are relatively pair-wise unrelated, e.g., group theory and PDEs) or depth (e.g., measure theory and functional analysis)?

Your question is philosophical rather than mathematical.

A colleague of mine told me the following metaphor / illustration once when I was a bachelor student and he did his PhD. And since now some years have passed I can relate.

It is hard to write it. Think about drawing a huge circle in the air, zooming in, and then drawing a huge circle again.

This is all knowledge:

[--------------------------------------------]

All knowledge contains a lot, and math is only a tiny part in it - marked with the cross:

[---------------------------------------x----]
                                        |
Zooming in:
[xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx]

Mathematical research is divided into many topics. Algebra, number theory, and many others, but also numerical mathematics. That is this tiny part here:

[xxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxxxxxx]
                    |
Zooming in:
[oooooooooooooooooooooooooooooooooooooooooooo]

Numerical Math is divided into several topics as well, like ODE numerics, optimisation etc. And one of them is FEM-Theory for PDEs.

[oooooooooooooooooooρoooooooooooooooooooooooo]
                    |

And that is the part of knowledge, where I feel comfortable saying "I know a bit more than most other people in the world".
Now after some years, I would extend that illustration one more step: My knowledge in that part rather looks like

[   ρ    ρρ  ρ         ρ   ρ          ρ     ρ]

I still only know "a bit" about it, most of it I don't know, and most of what I had learned is already forgotten.

(Actually FEM-Theory is still a huge topic, that contains e.g. different kinds of PDEs [elliptic, parabolic, hyperbolic, other]. So you could do the "zooming" several times more.)

Another small wisdom is:
Someone who finished school thinks he knows everything. Once he gained his masters degree, he knows that he knows nothing. And after the PhD he knows that  everyone around him knows nothing as well.

Asking about your focus: IMO use the first few years to explore topics in math to find out what you like. Then go deeper - if you found what you like.

Are there "must know" topics? There are basics that you learn in the first few terms. Without them it is hard to "speak" and "do" math. You will learn the tools that you need to dig deeper. After that feel free to enjoy math :)
If your research focus is for example on PDE numerics (as mine is) but you also like pure math - go ahead and take a lecture. Will it help you? Maybe, maybe not. But for sure you had fun gaining knowledge, and that is what counts.

Don't think too much about what lectures to attend. Everything will turn out all right. I think most mathematicians will agree with that statement.

Certainly not. For example, the great mathematician Grothendieck was insufficiently well acquainted with arithmetic to recognize the integer $57$ as a non-prime. The many accounts of this story can be accessed by an internet search for the key terms; say, look for grothendieck prime 57.

An earlier example is given by Ian Stewart on page 72 of his book Professor Stewart's Cabinet of Mathematical Curiosities, which I quote verbatim without any endorsement as to the degree of its truth:

Ernst Kummer was a German Algebraist, who did some of the best work on Fermat's Last Theorem before the modern era. However, he was poor at arithmetic, so he always asked his students to do the calculations for him. On one occasion he needed to work out $9\times7$. "Umm ... nine times seven is ... nine times ... seven ... is ..."

$\qquad$"Sixty-one," suggested one student. Kummer wrote this on the blackboard.

$\qquad$"No, Professor! It should be sixty-seven!" said another.

$\qquad$"Come, come, gentlemen," said Kummer. "It can't be both. It must be one or the other."